Question 1

In an equilibrium process,the concentrations of products and of reactants are equal.

Accepted Answer

In an equilibrium process, the rates of the forward and reverse reactions are equal, not necessarily the concentrations of products and reactants.

Question 2

Changes in temperature cause change in the equilibrium position.

Accepted Answer

According to Le Chatelier's principle, when a system at equilibrium is subjected to a stress, such as a change in temperature, pressure, or concentration, the system will shift in a way that reduces the stress. In the case of a change in temperature, if the reaction is exothermic, increasing the temperature will shift the equilibrium position to the left, favoring the reactants. On the other hand, if the reaction is endothermic, increasing the temperature will shift the equilibrium position to the right, favoring the products. Therefore, changes in temperature can cause a change in the equilibrium position.

Question 3

For which of the following reactions does K_p = K_c?

Accepted Answer

A

Question 4

What is the value for K_c for the reaction below if the equilibrium concentrations are [N₂] = 0.025 M,[H₂] = 0.0013 M and [NH₃] = 0.028 M for the following reaction? N₂(g)+ 3 H₂(g)⇌ 2 NH₃(g)

Accepted Answer

The equilibrium constant expression for the reaction is:K_c = [NH₃]²/[N₂][H₂]³Substituting the equilibrium concentrations into the expression, we get:K_c = (0.028 M)²/(0.025 M)(0.0013 M)³ = 1.4 × 10⁷

Question 5

The concentration of a pure solid is left out of a equilibrium constant expression but a pure liquid is included.

Accepted Answer

In a equilibrium constant expression, only the concentrations (or partial pressures) of the substances in the reaction mixture are included. Pure solids and liquids are not included because their concentrations do not change during the reaction. However, it is assumed that the activity (or effective concentration) of a pure liquid is constant and equal to 1, so it can be included in the equilibrium constant expression. Pure solids do not have an activity because they are not in the solution, so they are not included in the equilibrium constant expression.

Question 6

What is the value for K_c if [CO] = 0.025 M,[H₂] = 0.013 M and [CH₃OH] = 0.0028 M for the following reaction? CH₃OH(g)⇌ CO(g)+ 2 H₂(g)

Accepted Answer

The equilibrium constant expression for the given reaction is $K_c = \frac{[CO][H_2S]^2}{[CH_3OH]}$. Substituting the given concentrations, $K_c = \frac{(0.025)(0.013)^2}{0.0028} = 1.5 \times 10^{-3}$.

Question 7

Both products and reactants will be present in an equilibrium reaction unless K is very small or very large.

Accepted Answer

In an equilibrium reaction, both reactants and products are present in significant amounts unless the value of the equilibrium constant (K) is very small or very large. If K is very small, it means the reactants are favored and very few products will be formed, while if K is very large, it means the products are favored and very little reactants will be present.

Question 8

Which of the following substances present in the chemical reaction would be excluded from the equilibrium constant expression?

Accepted Answer

Only species that are in the same phase and participate in the reaction are included in the equilibrium constant expression. Therefore, the solvent (water) and any reactants that are not present in the same phase as the other reactants and products should be excluded. In this case, the liquid solvent (H₂O) should be excluded from the equilibrium constant expression.

Question 9

The value of the equilibrium constant for a given reaction depends on the initial concentrations of reactants.

Accepted Answer

The value of the equilibrium constant (Kc) remains constant at a given temperature, independent of the initial concentrations of reactants.

Question 10

In a system in equilibrium,two opposing reactions occur at equal rates.

Accepted Answer

In a system in equilibrium, the forward and reverse reactions occur at equal rates. This means that the rate of the forward reaction is equal to the rate of the reverse reaction. Therefore, both reactions are happening at the same speed, and they oppose each other to maintain the equilibrium state.

Question 11

Choose the INCORRECT statement.

Accepted Answer

A reversible chemical reaction is one in which equilibrium is established, not never established.

Question 12

Given that the equilibrium concentrations of [N₂] = 0.035 M,[C₂H₂] = 0.057 M,and [HCN] = 6.8 × 10^-4 M,find the value of the equilibrium constant expression for the reaction: N₂(g)+ C₂H₂(g)⇌ 2 HCN

Accepted Answer

The equilibrium constant expression for the reaction is given by: Kc = [HCN]2 / ([N₂] × [C₂H₂]) Substituting the given equilibrium concentrations: Kc = (6.8 × 10^-4)2 / (0.035 M × 0.057 M) ≈ 2.3 × 10^-4 Therefore, the best choice is D.

Question 13

For a solute (X)aq,in an ideal aqueous solution,its activity equals its concentration.

Accepted Answer

In an ideal aqueous solution, the activity of a solute is equal to its concentration. This is because an ideal solution follows Raoult's law, which states that the vapor pressure of each component in a solution is directly proportional to its mole fraction in the solution. In an ideal solution, the activity coefficient of a solute is equal to 1, which means that its activity is equal to its concentration.

Question 14

Given the following: I.N₂O(g)+ 1/2 O₂(g)⇌ 2 NO(g)K_c = 1.7 × 10^-13 II.N₂(g)+ O₂(g)⇌ 2 NO(g)K_c = 4.1 × 10^-31 Find the value of the equilibrium constant for the following equilibrium reaction: N₂(g)+ 1/2 O₂(g)⇌ N₂O(g)

Accepted Answer

We can calculate the equilibrium constant for the given reaction by multiplying the equilibrium constants of the two given reactions. 
Kc = (Kc1/2) * (Kc2)
Kc = (1.7 × 10^-13 / 2) * (4.1 × 10^-31)
Kc = 2.4 × 10^-18
Therefore, the correct answer is C.

Question 15

Large value for equilibrium constant K means the reaction will be less complete at the equilibrium point.

Accepted Answer

A large value for equilibrium constant K means that the products are favored and the reaction is more complete at equilibrium.

Question 16

One method to aid in working out equilibrium problems is the ICE table.

Accepted Answer

An ICE table is commonly used to assist with solving equilibrium problems.

Question 17

For the following chemical equilibrium,K_p = 4.6 × 10^-14 at 25 °C,find the value of K_c for this reaction at 25 °C. 2 Cl₂(g)+ 2 H₂O(g)⇌ 4 HCl(g)+ O₂(g)

Accepted Answer

The relationship between Kp and Kc is given by the equation Kp = Kc(RT)^(Δn), where Δn is the change in moles of gas. For this reaction, Δn = (4+1)-(2+2) = 1. At 25°C (298 K), R = 0.0821 L·atm/mol·K. Solving for Kc gives Kc = Kp / (RT)^(Δn) = 4.6 × 10^-14 / (0.0821 × 298)^1 = 1.9 × 10^-15.

Question 18

Reaction quotient Q will always be equal to the reaction's equilibrium constant K.

Accepted Answer

The reaction quotient Q is the ratio of the concentrations of products and reactants at any given point during a reaction. It can be calculated at any point during the reaction, whereas the equilibrium constant K is only valid when the reaction has reached equilibrium. Therefore, Q and K will only be equal when the reaction is at equilibrium.

Question 19

Equilibrium reactions are noted by a single straight arrow for a yield sign.

Accepted Answer

Equilibrium reactions are noted by a double arrow (⇌) indicating that the reaction can proceed in both forward and reverse directions. A single straight arrow is used for irreversible reactions where products are formed from reactants but the reaction cannot proceed in the reverse direction.

Question 20

For the reaction: H₂(g)+ I₂(g)⇌ 2 HI(g),K_c = 92.0 When equilibrium concentrations of HI and I₂ are [HI] = 0.115 M and [I₂] = 0.250 M,the equilibrium concentration of [H₂] is:

Accepted Answer

The equilibrium constant expression for the reaction is:K_c = [HI]²/[H₂][I₂]Substituting the given equilibrium concentrations into the expression, we get:92.0 = (0.115 M)²/[H₂](0.250 M)Solving for [H₂], we get:[H₂] = (0.115 M)²/(92.0 * 0.250 M) = 5.75 × 10^-4 MTherefore, the correct choice is B) 5.75 × 10^-4 M.

Quiz 15: Principles of Chemical Equilibrium

In an equilibrium process,the concentrations of products and of reactants are equal.

Changes in temperature cause change in the equilibrium position.

For which of the following reactions does K_p = K_c?

What is the value for K_c for the reaction below if the equilibrium concentrations are [N₂] = 0.025 M,[H₂] = 0.0013 M and [NH₃] = 0.028 M for the following reaction? N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g)

The concentration of a pure solid is left out of a equilibrium constant expression but a pure liquid is included.

What is the value for K_c if [CO] = 0.025 M,[H₂] = 0.013 M and [CH₃OH] = 0.0028 M for the following reaction? CH₃OH(g) ⇌ CO(g) + 2 H₂(g)

Both products and reactants will be present in an equilibrium reaction unless K is very small or very large.

Which of the following substances present in the chemical reaction would be excluded from the equilibrium constant expression?

The value of the equilibrium constant for a given reaction depends on the initial concentrations of reactants.

In a system in equilibrium,two opposing reactions occur at equal rates.

Choose the INCORRECT statement.

Given that the equilibrium concentrations of [N₂] = 0.035 M,[C₂H₂] = 0.057 M,and [HCN] = 6.8 × 10^-4 M,find the value of the equilibrium constant expression for the reaction: N₂(g) + C₂H₂(g) ⇌ 2 HCN

For a solute (X)aq,in an ideal aqueous solution,its activity equals its concentration.

Given the following: I.N₂O(g) + 1/2 O₂(g) ⇌ 2 NO(g) K_c = 1.7 × 10^-13 II.N₂(g) + O₂(g) ⇌ 2 NO(g) K_c = 4.1 × 10^-31 Find the value of the equilibrium constant for the following equilibrium reaction: N₂(g) + 1/2 O₂(g) ⇌ N₂O(g)

Large value for equilibrium constant K means the reaction will be less complete at the equilibrium point.

One method to aid in working out equilibrium problems is the ICE table.

For the following chemical equilibrium,K_p = 4.6 × 10^-14 at 25 °C,find the value of K_c for this reaction at 25 °C. 2 Cl₂(g) + 2 H₂O(g) ⇌ 4 HCl(g) + O₂(g)

Reaction quotient Q will always be equal to the reaction's equilibrium constant K.

Equilibrium reactions are noted by a single straight arrow for a yield sign.

For the reaction: H₂(g) + I₂(g) ⇌ 2 HI(g) ,K_c = 92.0 When equilibrium concentrations of HI and I₂ are [HI] = 0.115 M and [I₂] = 0.250 M,the equilibrium concentration of [H₂] is:

Quiz 15: Principles of Chemical Equilibrium

In an equilibrium process,the concentrations of products and of reactants are equal.

Changes in temperature cause change in the equilibrium position.

For which of the following reactions does Kp = Kc?

What is the value for Kc for the reaction below if the equilibrium concentrations are [N2] = 0.025 M,[H2] = 0.0013 M and [NH3] = 0.028 M for the following reaction? N2(g) + 3 H2(g) ⇌ 2 NH3(g)

The concentration of a pure solid is left out of a equilibrium constant expression but a pure liquid is included.

What is the value for Kc if [CO] = 0.025 M,[H2] = 0.013 M and [CH3OH] = 0.0028 M for the following reaction? CH3OH(g) ⇌ CO(g) + 2 H2(g)

Both products and reactants will be present in an equilibrium reaction unless K is very small or very large.

Which of the following substances present in the chemical reaction would be excluded from the equilibrium constant expression?

The value of the equilibrium constant for a given reaction depends on the initial concentrations of reactants.

In a system in equilibrium,two opposing reactions occur at equal rates.

Choose the INCORRECT statement.

Given that the equilibrium concentrations of [N2] = 0.035 M,[C2H2] = 0.057 M,and [HCN] = 6.8 × 10-4 M,find the value of the equilibrium constant expression for the reaction: N2(g) + C2H2(g) ⇌ 2 HCN

For a solute (X)aq,in an ideal aqueous solution,its activity equals its concentration.

Given the following: I.N2O(g) + 1/2 O2(g) ⇌ 2 NO(g) Kc = 1.7 × 10-13 II.N2(g) + O2(g) ⇌ 2 NO(g) Kc = 4.1 × 10-31 Find the value of the equilibrium constant for the following equilibrium reaction: N2(g) + 1/2 O2(g) ⇌ N2O(g)

Large value for equilibrium constant K means the reaction will be less complete at the equilibrium point.

One method to aid in working out equilibrium problems is the ICE table.

For the following chemical equilibrium,Kp = 4.6 × 10-14 at 25 °C,find the value of Kc for this reaction at 25 °C. 2 Cl2(g) + 2 H2O(g) ⇌ 4 HCl(g) + O2(g)

Reaction quotient Q will always be equal to the reaction's equilibrium constant K.

Equilibrium reactions are noted by a single straight arrow for a yield sign.

For the reaction: H2(g) + I2(g) ⇌ 2 HI(g) ,Kc = 92.0 When equilibrium concentrations of HI and I2 are [HI] = 0.115 M and [I2] = 0.250 M,the equilibrium concentration of [H2] is:

For which of the following reactions does K_p = K_c?

What is the value for K_c for the reaction below if the equilibrium concentrations are [N₂] = 0.025 M,[H₂] = 0.0013 M and [NH₃] = 0.028 M for the following reaction? N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g)

What is the value for K_c if [CO] = 0.025 M,[H₂] = 0.013 M and [CH₃OH] = 0.0028 M for the following reaction? CH₃OH(g) ⇌ CO(g) + 2 H₂(g)

Given that the equilibrium concentrations of [N₂] = 0.035 M,[C₂H₂] = 0.057 M,and [HCN] = 6.8 × 10^-4 M,find the value of the equilibrium constant expression for the reaction: N₂(g) + C₂H₂(g) ⇌ 2 HCN

Given the following: I.N₂O(g) + 1/2 O₂(g) ⇌ 2 NO(g) K_c = 1.7 × 10^-13 II.N₂(g) + O₂(g) ⇌ 2 NO(g) K_c = 4.1 × 10^-31 Find the value of the equilibrium constant for the following equilibrium reaction: N₂(g) + 1/2 O₂(g) ⇌ N₂O(g)

For the following chemical equilibrium,K_p = 4.6 × 10^-14 at 25 °C,find the value of K_c for this reaction at 25 °C. 2 Cl₂(g) + 2 H₂O(g) ⇌ 4 HCl(g) + O₂(g)

For the reaction: H₂(g) + I₂(g) ⇌ 2 HI(g) ,K_c = 92.0 When equilibrium concentrations of HI and I₂ are [HI] = 0.115 M and [I₂] = 0.250 M,the equilibrium concentration of [H₂] is: