Question 1

Consider the equilibrium system: N₂O₄(g)⇌ 2 NO₂(g)for which K_p = 0.1134 at 25 °C and Δ_rH° = 58.03 kJ/mol.Assuming that the total pressure inside the container is 1.00 atm at equilibrium and that initially only N₂O₄ was present inside the container,compute At equilibrium.

Accepted Answer

Using the ideal gas law, we can express the equilibrium partial pressures in terms of the equilibrium mole fractions. Let x be the mole fraction of NO₂(g) at equilibrium. Then, the equilibrium mole fractions are: [NO₂(g)] = 2x [N₂O₄] = 1 - 2x Using the equilibrium constant expression: K_p = [NO₂(g)]^2/[N₂O₄] = (2x)^2/(1-2x) At equilibrium, the partial pressure of NO₂(g) is 2xPtotal, where Ptotal = 1.00 atm. Similarly, the partial pressure of N₂O₄(g) is (1-2x)Ptotal. Using the ideal gas law, we can convert partial pressures to concentrations: [NO₂(g)] = 2xPtotal/RT [N₂O₄] = (1-2x)Ptotal/RT Substituting into the equilibrium constant expression and solving for x: 0.1134 = (2xPtotal/RT)^2/[(1-2x)Ptotal/RT] 0.1134 = 4x^2/(1-2x) 0.1134 - 0.1134x + 4x^2 = 0 (2x - 0.285)(1-14x) = 0 The positive root, x = 0.142, corresponds to the mole fraction of NO₂(g) at equilibrium. Therefore, the partial pressures at equilibrium are: [NO₂(g)] = 2xPtotal = 0.285 atm [N₂O₄] = (1-2x)Ptotal = 0.715 atm Thus, the answer is D) 0.285 atm.

Question 2

Consider the following reaction: POCl₃(g)⇌ POCl(g)+ Cl₂(g)with K_c = 0.450 A sample of pure POCl₃(g)was placed in a reaction vessel and allowed to decompose according to the above reaction.At equilibrium,the concentrations of POCl(g)and Cl₂(g)were each 0.150 M.What was the initial concentration of POCl₃(g)?

Accepted Answer

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Question 3

A mixture,containing 0.0750 M HCl(g)and 0.0330 M O₂(g)is allowed to come to equilibrium at 480 °C. 4 HCl(g)+ O₂(g)⇌ 2 Cl₂(g)+ 2 H₂O(g) At equilibrium [Cl₂] = 0.030 M.What is the value of K_c?

Accepted Answer

From the balanced equation, we can see that the equilibrium constant expression is: K_c = ([Cl₂]2[H₂O]) / ([HCl]4[O₂]) Substituting the given values at equilibrium: 0.0302 = (0.030)2(2x) / (0.0750)4(0.0330) Simplifying and solving for x: x = [H₂O] = 8.90 × 10^-3 M Substituting all values into the equilibrium constant expression: K_c = (0.0302)(2x) / (0.0750)4(0.0330) = 890 Therefore, the answer is B.

Question 4

A mixture containing 0.392 M A(g)and 0.452 M B(g)is allowed to come to equilibrium at 300 K.The reaction 3 A(g)+ 2 B(g)⇌ C(g)+ D(g)occurs.At equilibrium,[C] = 0.00128 M.What is the value of K_c?

Accepted Answer

To find Kc, use the equilibrium expression: Kc = [C][D] / ([A]^3[B]^2). At equilibrium, [C] = 0.00128 M, and since the stoichiometry is 1:1 for C and D, [D] = 0.00128 M. The change in concentration for A and B can be calculated using stoichiometry, and then substituted into the expression to find Kc.

Question 5

For the reaction: PCl₃(g)+ Cl₂(g)⇌ PCl₅(g)at 85 °C,K_p = 1.19 If one starts with 2.00 atm pressure of PCl₃,1.00 atm pressure of Cl₂ and no PCl₅,what is the partial pressure of PCl₅(g)at equilibrium?

Accepted Answer

To find the partial pressure of PCl5 at equilibrium, set up the expression for Kp: Kp = (P_PCl5) / (P_PCl3 * P_Cl2). Let x be the change in pressure for PCl5, then at equilibrium: P_PCl5 = x, P_PCl3 = 2.00 - x, P_Cl2 = 1.00 - x. Substitute these into the Kp expression and solve for x: 1.19 = x / ((2.00 - x)(1.00 - x)). Solving this equation gives x ≈ 0.621 atm.

Question 6

Consider the equilibrium system: N₂O₄(g)⇌ 2 NO₂(g)for which K_p = 0.1134 at 25 °C and Δ_rH° = 58.03 kJ/mol.Assuming that the total pressure inside the container is 1.00 atm at equilibrium and that initially only N₂O₄ was present inside the container,compute At equilibrium.

Accepted Answer

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Question 7

Consider the equilibrium system: N₂O₄(g)⇌ 2 NO₂(g)for which K_p = 0.1134 at 25 °C and Δ_rH° = 58.03 kJ/mol.Assuming that the total pressure inside the container is 10 atm at equilibrium and that initially only N₂O₄ was present inside the container,compute At equilibrium.

Accepted Answer

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Question 8

At 35 °C,K_p = 0.315 for the reaction N₂O₄(g)⇌ 2 NO₂(g),if the initial pressure of NO₂(g)in a container is 3.00 atm,what is the equilibrium pressure of N₂O₄(g)?

Accepted Answer

To find the equilibrium pressure of N₂O₄, we set up the expression for Kp: Kp = (P_NO₂)² / (P_N₂O₄). Let x be the change in pressure for N₂O₄, then at equilibrium, P_NO₂ = 3.00 - 2x and P_N₂O₄ = x. Solving the equation 0.315 = ((3.00 - 2x)²) / x gives x = 1.19 atm.

Question 9

For the decomposition of ammonium carbamate NH₄(NH₂CO₂)(s)⇌ 2 NH₃(g)+ CO₂(g) K_p = 0.0596 at a certain temperature.A solid sample of ammonium carbamate is introduced into an evacuated container and at equilibrium some solid remains in the container.What is the total pressure in the container?

Accepted Answer

The total pressure in the container can be calculated using the equilibrium constant (Kp) for the decomposition reaction. Given that Kp = 0.0596, and the reaction produces 2 moles of NH3 and 1 mole of CO2 for every mole of ammonium carbamate that decomposes, the total pressure is determined by the partial pressures of NH3 and CO2 at equilibrium. Since the reaction produces 3 moles of gas in total and some solid remains, indicating the reaction has not gone to completion, the equilibrium condition can be represented as Kp = (P_NH3)^2 * P_CO2. Given the stoichiometry of the reaction, P_NH3 is twice that of P_CO2 at equilibrium. By solving the equilibrium expression for the total pressure (P_total = 2P_NH3 + P_CO2) and using the value of Kp, the total pressure is calculated to be approximately 0.738 atm, assuming ideal behavior of the gases involved.

Question 10

Consider the following reversible reaction: POCl₃(g)⇌ POCl(g)+ Cl₂(g)K_c = 0.450 The following initial amounts of reactants and products were mixed: [POCl₃] = 0.750 M,[POCl] = 0.550 M,and [Cl₂] = 0.150 M.What is the equilibrium concentration of POCl?

Accepted Answer

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Question 11

4.000 mol chlorine and 2.000 mol bromine were placed in a 50.0 L container and kept at 293 K until equilibrium was achieved for the reaction: Br₂(l)+ Cl₂(g)⇌ 2 BrCl(g) At the point of equilibrium there were 82.63 g of Br₂(l).Determine the total pressure in the 50.0 L container at equilibrium.

Accepted Answer

To find the total pressure, first calculate the moles of Br2 at equilibrium using its mass (82.63 g) and molar mass (159.808 g/mol), which gives approximately 0.517 mol of Br2. The change in moles of Br2 from initial to equilibrium is 2.000 - 0.517 = 1.483 mol, which means 1.483 mol of Cl2 reacted, and 2 * 1.483 = 2.966 mol of BrCl was formed. The total moles of gas at equilibrium is 4.000 - 1.483 + 2.966 = 5.483 mol. Using the ideal gas law (PV=nRT) with R = 0.0821 L·atm/mol·K and T = 293 K, the total pressure is calculated as (5.483 mol * 0.0821 L·atm/mol·K * 293 K) / 50.0 L = 2.64 atm.

Question 12

Consider the following equilibrium. A(g)+ 3 B(g)⇌ 2 C(g) If the initial concentrations are [A] = 1.00 M,[B] = 3.00 M,and [C] = 0,at equilibrium it is found that [C] = 0.980 M.Calculate K_c for this reaction.

Accepted Answer

The equilibrium constant $K_c$ for the reaction $A(g) + 3B(g) \rightleftharpoons 2C(g)$ can be calculated using the equilibrium concentrations of the reactants and products. Given that at equilibrium, $[C] = 0.980 \, M$, we can determine the changes in concentrations of $A$ and $B$ based on the stoichiometry of the reaction. For every 2 moles of $C$ produced, 1 mole of $A$ and 3 moles of $B$ are consumed.Since $[C]$ increases by $0.980 \, M$, the change in $[A]$ is $-0.490 \, M$ (half of the change in $[C]$) and the change in $[B]$ is $-3 \times 0.490 \, M = -1.470 \, M$. Therefore, the equilibrium concentrations are:- $[A]_{eq} = 1.00 \, M - 0.490 \, M = 0.510 \, M$- $[B]_{eq} = 3.00 \, M - 1.470 \, M = 1.530 \, M$- $[C]_{eq} = 0.980 \, M$The equilibrium constant $K_c$ is given by:$$K_c = \frac{[C]^2}{[A][B]^3}$$Plugging in the equilibrium concentrations:$$K_c = \frac{(0.980)^2}{(0.510)(1.530)^3} = \frac{0.9604}{0.510 \times 3.5809} = \frac{0.9604}{1.8276} \approx 0.526$$Therefore, the correct answer is $0.526$, which corresponds to choice A.

Question 13

A mixture containing 0.0392 M A(g)and 0.0452 M B(g)is allowed to come to equilibrium at 300 K.The reaction: 3 A(g)+ 2 B(g)⇌ C(g)+ D(g)occurs.At equilibrium,[C] = 0.00128 M.What is the value of K_c?

Accepted Answer

The answer of A mixture containing 0.0392 M A(g)and 0.0452...

Question 14

One of the reactions in the gasification of coal is CO(g)+ 3 H₂(g)⇌ CH₄(g)+ H₂O(g)Δ_rH° = -237 kJ For each of the following changes,explain the effect (increase,decrease,no effect)on the number of moles of CH₄(g)produced. (a)increase the concentration of H₂O(g) (b)decrease the volume of the container (c)add a catalyst (d)decrease the temperature

Accepted Answer

The answer of One of the reactions in the gasification...

Question 15

Consider the equilibrium system: N₂O₄(g)⇌ 2 NO₂(g)for which K_p = 0.1134 at 25 °C and Δ_rH° = 58.03 kJ/mol.Assume that 1 mole of N₂O₄ and 2 moles of NO₂ are introduced into a 5.0 liter container.What will be the equilibrium value of [N₂O]?

Accepted Answer

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Question 16

4.000 mol chlorine and 2.000 mol bromine were placed in a 50.0 L container and kept at 293 K until equilibrium was achieved for the reaction: Br₂(l)+ Cl₂(g)⇌ BrCl(g) At the point of equilibrium there were 82.63 g of Br₂(l).Compute the value of K_c for this reaction.

Accepted Answer

To calculate the equilibrium constant $K_c$, we first need to determine the equilibrium concentrations of the reactants and products. Given that 82.63 g of $Br_2$ are present at equilibrium, we can calculate the moles of $Br_2$ left:Molar mass of $Br_2 = 79.904 \times 2 = 159.808 \, \text{g/mol}$Moles of $Br_2$ at equilibrium = $\frac{82.63 \, \text{g}}{159.808 \, \text{g/mol}} = 0.517 \, \text{mol}$Since we started with 2.000 mol of $Br_2$, the change in moles of $Br_2$ (and thus $Cl_2$, since they react in a 1:1 ratio) is $2.000 - 0.517 = 1.483 \, \text{mol}$. This is also the amount of $BrCl$ formed, as it is produced from $Br_2$ and $Cl_2$ in a 1:1 ratio.The equilibrium concentrations are then:- $[Br_2] = \frac{0.517 \, \text{mol}}{50.0 \, \text{L}} = 0.01034 \, \text{M}$- $[Cl_2] = \frac{4.000 \, \text{mol} - 1.483 \, \text{mol}}{50.0 \, \text{L}} = 0.05034 \, \text{M}$ (since we started with 4.000 mol of $Cl_2$)- $[BrCl] = \frac{1.483 \, \text{mol}}{50.0 \, \text{L}} = 0.02966 \, \text{M}$The equilibrium constant $K_c$ is given by:$$K_c = \frac{[BrCl]}{[Br_2][Cl_2]} = \frac{0.02966}{0.01034 \times 0.05034} = 0.0699$$Therefore, the correct answer is A) 0.0699.

Question 17

Consider the equilibrium system: N₂O₄(g)⇌ 2 NO₂(g)for which K_p = 0.1134 at 25 °C and Δ_rH° = 58.03 kJ/mol.Assuming that the total pressure inside the container is 10 atm at equilibrium and that initially only N₂O₄ was present inside the container,compute At equilibrium.

Accepted Answer

The answer of Consider the equilibrium system: N₂O₄(g)⇌ 2 NO₂(g)for...

Question 18

Consider the equilibrium system: N₂O₄(g)⇌ 2 NO₂(g)for which K_p = 0.1134 at 25 °C and Δ_rH° = 58.03 kJ/mol.Assume that 1 mole of N₂O₄ and 2 moles of NO₂ are introduced into a 5.0 liter container.What will be the equilibrium value of [N₂O₄]?

Accepted Answer

We can use the equation: ΔG° = -RTln(K) where ΔG° is the standard free energy change, R is the gas constant, T is the temperature in Kelvin, and ln(K) is the natural logarithm of the equilibrium constant. First, we need to calculate ΔG° using the equation: ΔG° = ΔH° - TΔS° where ΔH° is the standard enthalpy change, ΔS° is the standard entropy change, and T is the temperature in Kelvin. We are given ΔS° = 58.03 kJ/mol and we can assume that ΔH° is independent of temperature. Therefore, we can calculate ΔG° at 25 °C (298 K): ΔG° = (58.03 kJ/mol) - (298 K)(0.05803 kJ/mol K) = 40.67 kJ/mol Next, we can use the equation above to calculate ln(K): ln(K) = -ΔG°/RT = -(40.67 kJ/mol)/[(8.314 J/mol K)(298 K)] = -15.41 To solve for the equilibrium concentration of N₂O₄, we can use the equilibrium expression: K = [NO₂]2/[N₂O₄] And since we know that [NO₂]/[N₂O₄] = 2 (because the stoichiometric coefficient of NO₂ is 2 in the balanced equation), we can substitute to get: K = 2²/[N₂O₄] = 4/[N₂O₄] Solving for [N₂O₄], we get: [N₂O₄] = 4/K = 4/exp(-15.41) = 0.379 M

Question 19

Consider the following reaction occurring at 960 K: CO₂(g)+ H₂(g)⇌ CO(g)+ H₂O(g) At equilibrium,the concentrations of reactants and products are: [CO₂] = 0.0400 M,[H₂] = 0.0220 M,[CO] = 0.0240 M,and [H₂O] = 0.0190 M.This equilibrium is perturbed by adding CO₂(g)to the system such that when a new equilibrium is reached,the concentration of CO₂ is 0.050 M.What is the concentration of H₂(g)at this equilibrium?

Accepted Answer

The addition of COS (carbon oxysulfide) shifts the equilibrium to the right according to Le Chatelier's principle, increasing the consumption of H2S (hydrogen sulfide) and the production of CO (carbon monoxide) and H2SO (hydrogen thioperoxide). Since the initial and final concentrations of COS are given, and assuming the reaction proceeds without any side reactions, the change in concentration of COS (from 0.0400 M to 0.0500 M) directly affects the concentration of H2S. However, without the stoichiometry of the reaction or additional information, the exact change in H2S concentration cannot be calculated precisely from the given data. The correct answer, therefore, is based on understanding the qualitative effect of the reaction shift rather than a quantitative calculation that can be directly inferred from the provided information.

Question 20

For the reaction: PCl₃(g)+ Cl₂(g)⇌ PCl₅(g)at 70.5 °C,K_p = 1.05. If one starts with 1.80 atm pressure of PCl₃(g),1.72 atm pressure of Cl₂(g),and no PCl₅(g),what is the partial pressure of PCl₅(g)at equilibrium?

Accepted Answer

The equilibrium constant expression for the reaction is $K_p = \frac{[PCl_5]}{[PCl_3][Cl_2]}$. Given $K_p = 1.05$, initial pressures of $PCl_3 = 1.80$ atm and $Cl_2 = 1.72$ atm, and $PCl_5 = 0$ atm, we can set up an ICE table. At equilibrium, the pressures change by $x$: $PCl_3$ and $Cl_2$ decrease by $x$, and $PCl_5$ increases by $x$. Thus, $K_p = \frac{x}{(1.80-x)(1.72-x)} = 1.05$. Solving this equation for $x$ gives the change in concentration, and since $PCl_5$ starts at 0 atm, its equilibrium pressure is equal to $x$, which is approximately 0.856 atm.

Quiz 15: Principles of Chemical Equilibrium

Consider the equilibrium system: N₂O₄(g) ⇌ 2 NO₂(g) for which K_p = 0.1134 at 25 °C and Δ_rH° = 58.03 kJ/mol.Assuming that the total pressure inside the container is 1.00 atm at equilibrium and that initially only N₂O₄ was present inside the container,compute At equilibrium.

A mixture,containing 0.0750 M HCl(g) and 0.0330 M O₂(g) is allowed to come to equilibrium at 480 °C. 4 HCl(g) + O₂(g) ⇌ 2 Cl₂(g) + 2 H₂O(g) At equilibrium [Cl₂] = 0.030 M.What is the value of K_c?

A mixture containing 0.392 M A(g) and 0.452 M B(g) is allowed to come to equilibrium at 300 K.The reaction 3 A(g) + 2 B(g) ⇌ C(g) + D(g) occurs.At equilibrium,[C] = 0.00128 M.What is the value of K_c?

For the reaction: PCl₃(g) + Cl₂(g) ⇌ PCl₅(g) at 85 °C,K_p = 1.19 If one starts with 2.00 atm pressure of PCl₃,1.00 atm pressure of Cl₂ and no PCl₅,what is the partial pressure of PCl₅(g) at equilibrium?

Consider the equilibrium system: N₂O₄(g) ⇌ 2 NO₂(g) for which K_p = 0.1134 at 25 °C and Δ_rH° = 58.03 kJ/mol.Assuming that the total pressure inside the container is 1.00 atm at equilibrium and that initially only N₂O₄ was present inside the container,compute At equilibrium.

Consider the equilibrium system: N₂O₄(g) ⇌ 2 NO₂(g) for which K_p = 0.1134 at 25 °C and Δ_rH° = 58.03 kJ/mol.Assuming that the total pressure inside the container is 10 atm at equilibrium and that initially only N₂O₄ was present inside the container,compute At equilibrium.

At 35 °C,K_p = 0.315 for the reaction N₂O₄(g) ⇌ 2 NO₂(g) ,if the initial pressure of NO₂(g) in a container is 3.00 atm,what is the equilibrium pressure of N₂O₄(g) ?

Consider the following reversible reaction: POCl₃(g) ⇌ POCl(g) + Cl₂(g) K_c = 0.450 The following initial amounts of reactants and products were mixed: [POCl₃] = 0.750 M,[POCl] = 0.550 M,and [Cl₂] = 0.150 M.What is the equilibrium concentration of POCl?

Consider the following equilibrium. A(g) + 3 B(g) ⇌ 2 C(g) If the initial concentrations are [A] = 1.00 M,[B] = 3.00 M,and [C] = 0,at equilibrium it is found that [C] = 0.980 M.Calculate K_c for this reaction.

A mixture containing 0.0392 M A(g) and 0.0452 M B(g) is allowed to come to equilibrium at 300 K.The reaction: 3 A(g) + 2 B(g) ⇌ C(g) + D(g) occurs.At equilibrium,[C] = 0.00128 M.What is the value of K_c?

Consider the equilibrium system: N₂O₄(g) ⇌ 2 NO₂(g) for which K_p = 0.1134 at 25 °C and Δ_rH° = 58.03 kJ/mol.Assume that 1 mole of N₂O₄ and 2 moles of NO₂ are introduced into a 5.0 liter container.What will be the equilibrium value of [N₂O]?

4.000 mol chlorine and 2.000 mol bromine were placed in a 50.0 L container and kept at 293 K until equilibrium was achieved for the reaction: Br₂(l) + Cl₂(g) ⇌ BrCl(g) At the point of equilibrium there were 82.63 g of Br₂(l) .Compute the value of K_c for this reaction.

Consider the equilibrium system: N₂O₄(g) ⇌ 2 NO₂(g) for which K_p = 0.1134 at 25 °C and Δ_rH° = 58.03 kJ/mol.Assuming that the total pressure inside the container is 10 atm at equilibrium and that initially only N₂O₄ was present inside the container,compute At equilibrium.

Consider the equilibrium system: N₂O₄(g) ⇌ 2 NO₂(g) for which K_p = 0.1134 at 25 °C and Δ_rH° = 58.03 kJ/mol.Assume that 1 mole of N₂O₄ and 2 moles of NO₂ are introduced into a 5.0 liter container.What will be the equilibrium value of [N₂O₄]?

For the reaction: PCl₃(g) + Cl₂(g) ⇌ PCl₅(g) at 70.5 °C,K_p = 1.05. If one starts with 1.80 atm pressure of PCl₃(g) ,1.72 atm pressure of Cl₂(g) ,and no PCl₅(g) ,what is the partial pressure of PCl₅(g) at equilibrium?

Quiz 15: Principles of Chemical Equilibrium

Consider the equilibrium system: N2O4(g) ⇌ 2 NO2(g) for which Kp = 0.1134 at 25 °C and ΔrH° = 58.03 kJ/mol.Assuming that the total pressure inside the container is 1.00 atm at equilibrium and that initially only N2O4 was present inside the container,compute At equilibrium.

A mixture,containing 0.0750 M HCl(g) and 0.0330 M O2(g) is allowed to come to equilibrium at 480 °C. 4 HCl(g) + O2(g) ⇌ 2 Cl2(g) + 2 H2O(g) At equilibrium [Cl2] = 0.030 M.What is the value of Kc?

A mixture containing 0.392 M A(g) and 0.452 M B(g) is allowed to come to equilibrium at 300 K.The reaction 3 A(g) + 2 B(g) ⇌ C(g) + D(g) occurs.At equilibrium,[C] = 0.00128 M.What is the value of Kc?

For the reaction: PCl3(g) + Cl2(g) ⇌ PCl5(g) at 85 °C,Kp = 1.19 If one starts with 2.00 atm pressure of PCl3,1.00 atm pressure of Cl2 and no PCl5,what is the partial pressure of PCl5(g) at equilibrium?

Consider the equilibrium system: N2O4(g) ⇌ 2 NO2(g) for which Kp = 0.1134 at 25 °C and ΔrH° = 58.03 kJ/mol.Assuming that the total pressure inside the container is 1.00 atm at equilibrium and that initially only N2O4 was present inside the container,compute At equilibrium.

Consider the equilibrium system: N2O4(g) ⇌ 2 NO2(g) for which Kp = 0.1134 at 25 °C and ΔrH° = 58.03 kJ/mol.Assuming that the total pressure inside the container is 10 atm at equilibrium and that initially only N2O4 was present inside the container,compute At equilibrium.

At 35 °C,Kp = 0.315 for the reaction N2O4(g) ⇌ 2 NO2(g) ,if the initial pressure of NO2(g) in a container is 3.00 atm,what is the equilibrium pressure of N2O4(g) ?

For the decomposition of ammonium carbamate NH4(NH2CO2) (s) ⇌ 2 NH3(g) + CO2(g) Kp = 0.0596 at a certain temperature.A solid sample of ammonium carbamate is introduced into an evacuated container and at equilibrium some solid remains in the container.What is the total pressure in the container?

Consider the following reversible reaction: POCl3(g) ⇌ POCl(g) + Cl2(g) Kc = 0.450 The following initial amounts of reactants and products were mixed: [POCl3] = 0.750 M,[POCl] = 0.550 M,and [Cl2] = 0.150 M.What is the equilibrium concentration of POCl?

4.000 mol chlorine and 2.000 mol bromine were placed in a 50.0 L container and kept at 293 K until equilibrium was achieved for the reaction: Br2(l) + Cl2(g) ⇌ 2 BrCl(g) At the point of equilibrium there were 82.63 g of Br2(l) .Determine the total pressure in the 50.0 L container at equilibrium.

Consider the following equilibrium. A(g) + 3 B(g) ⇌ 2 C(g) If the initial concentrations are [A] = 1.00 M,[B] = 3.00 M,and [C] = 0,at equilibrium it is found that [C] = 0.980 M.Calculate Kc for this reaction.

A mixture containing 0.0392 M A(g) and 0.0452 M B(g) is allowed to come to equilibrium at 300 K.The reaction: 3 A(g) + 2 B(g) ⇌ C(g) + D(g) occurs.At equilibrium,[C] = 0.00128 M.What is the value of Kc?

Consider the equilibrium system: N2O4(g) ⇌ 2 NO2(g) for which Kp = 0.1134 at 25 °C and ΔrH° = 58.03 kJ/mol.Assume that 1 mole of N2O4 and 2 moles of NO2 are introduced into a 5.0 liter container.What will be the equilibrium value of [N2O]?

4.000 mol chlorine and 2.000 mol bromine were placed in a 50.0 L container and kept at 293 K until equilibrium was achieved for the reaction: Br2(l) + Cl2(g) ⇌ BrCl(g) At the point of equilibrium there were 82.63 g of Br2(l) .Compute the value of Kc for this reaction.

Consider the equilibrium system: N2O4(g) ⇌ 2 NO2(g) for which Kp = 0.1134 at 25 °C and ΔrH° = 58.03 kJ/mol.Assuming that the total pressure inside the container is 10 atm at equilibrium and that initially only N2O4 was present inside the container,compute At equilibrium.

Consider the equilibrium system: N2O4(g) ⇌ 2 NO2(g) for which Kp = 0.1134 at 25 °C and ΔrH° = 58.03 kJ/mol.Assume that 1 mole of N2O4 and 2 moles of NO2 are introduced into a 5.0 liter container.What will be the equilibrium value of [N2O4]?

For the reaction: PCl3(g) + Cl2(g) ⇌ PCl5(g) at 70.5 °C,Kp = 1.05. If one starts with 1.80 atm pressure of PCl3(g) ,1.72 atm pressure of Cl2(g) ,and no PCl5(g) ,what is the partial pressure of PCl5(g) at equilibrium?

Consider the equilibrium system: N₂O₄(g) ⇌ 2 NO₂(g) for which K_p = 0.1134 at 25 °C and Δ_rH° = 58.03 kJ/mol.Assuming that the total pressure inside the container is 1.00 atm at equilibrium and that initially only N₂O₄ was present inside the container,compute At equilibrium.

A mixture,containing 0.0750 M HCl(g) and 0.0330 M O₂(g) is allowed to come to equilibrium at 480 °C. 4 HCl(g) + O₂(g) ⇌ 2 Cl₂(g) + 2 H₂O(g) At equilibrium [Cl₂] = 0.030 M.What is the value of K_c?

A mixture containing 0.392 M A(g) and 0.452 M B(g) is allowed to come to equilibrium at 300 K.The reaction 3 A(g) + 2 B(g) ⇌ C(g) + D(g) occurs.At equilibrium,[C] = 0.00128 M.What is the value of K_c?

For the reaction: PCl₃(g) + Cl₂(g) ⇌ PCl₅(g) at 85 °C,K_p = 1.19 If one starts with 2.00 atm pressure of PCl₃,1.00 atm pressure of Cl₂ and no PCl₅,what is the partial pressure of PCl₅(g) at equilibrium?

Consider the equilibrium system: N₂O₄(g) ⇌ 2 NO₂(g) for which K_p = 0.1134 at 25 °C and Δ_rH° = 58.03 kJ/mol.Assuming that the total pressure inside the container is 1.00 atm at equilibrium and that initially only N₂O₄ was present inside the container,compute At equilibrium.

Consider the equilibrium system: N₂O₄(g) ⇌ 2 NO₂(g) for which K_p = 0.1134 at 25 °C and Δ_rH° = 58.03 kJ/mol.Assuming that the total pressure inside the container is 10 atm at equilibrium and that initially only N₂O₄ was present inside the container,compute At equilibrium.

At 35 °C,K_p = 0.315 for the reaction N₂O₄(g) ⇌ 2 NO₂(g) ,if the initial pressure of NO₂(g) in a container is 3.00 atm,what is the equilibrium pressure of N₂O₄(g) ?

Consider the following reversible reaction: POCl₃(g) ⇌ POCl(g) + Cl₂(g) K_c = 0.450 The following initial amounts of reactants and products were mixed: [POCl₃] = 0.750 M,[POCl] = 0.550 M,and [Cl₂] = 0.150 M.What is the equilibrium concentration of POCl?

Consider the following equilibrium. A(g) + 3 B(g) ⇌ 2 C(g) If the initial concentrations are [A] = 1.00 M,[B] = 3.00 M,and [C] = 0,at equilibrium it is found that [C] = 0.980 M.Calculate K_c for this reaction.

A mixture containing 0.0392 M A(g) and 0.0452 M B(g) is allowed to come to equilibrium at 300 K.The reaction: 3 A(g) + 2 B(g) ⇌ C(g) + D(g) occurs.At equilibrium,[C] = 0.00128 M.What is the value of K_c?

Consider the equilibrium system: N₂O₄(g) ⇌ 2 NO₂(g) for which K_p = 0.1134 at 25 °C and Δ_rH° = 58.03 kJ/mol.Assume that 1 mole of N₂O₄ and 2 moles of NO₂ are introduced into a 5.0 liter container.What will be the equilibrium value of [N₂O]?

4.000 mol chlorine and 2.000 mol bromine were placed in a 50.0 L container and kept at 293 K until equilibrium was achieved for the reaction: Br₂(l) + Cl₂(g) ⇌ BrCl(g) At the point of equilibrium there were 82.63 g of Br₂(l) .Compute the value of K_c for this reaction.

Consider the equilibrium system: N₂O₄(g) ⇌ 2 NO₂(g) for which K_p = 0.1134 at 25 °C and Δ_rH° = 58.03 kJ/mol.Assuming that the total pressure inside the container is 10 atm at equilibrium and that initially only N₂O₄ was present inside the container,compute At equilibrium.

Consider the equilibrium system: N₂O₄(g) ⇌ 2 NO₂(g) for which K_p = 0.1134 at 25 °C and Δ_rH° = 58.03 kJ/mol.Assume that 1 mole of N₂O₄ and 2 moles of NO₂ are introduced into a 5.0 liter container.What will be the equilibrium value of [N₂O₄]?

For the reaction: PCl₃(g) + Cl₂(g) ⇌ PCl₅(g) at 70.5 °C,K_p = 1.05. If one starts with 1.80 atm pressure of PCl₃(g) ,1.72 atm pressure of Cl₂(g) ,and no PCl₅(g) ,what is the partial pressure of PCl₅(g) at equilibrium?