Question 1

For the decomposition of calcium carbonate,ΔG° = +130.4 kJ at 25°C. CaCO₃(s)→ CaO(s)+ CO₂(g) Calculate the partial pressure of CO₂ if ΔG = 0.0 kJ.(R = 8.31 × 10−³ kJ/K)

Accepted Answer

The correct calculation involves using the equation ΔG° = -RTlnKp, where ΔG° is the standard free energy change, R is the gas constant, T is the temperature in Kelvin, and Kp is the equilibrium constant related to the partial pressures of the gases involved. Given ΔG° = +130.4 kJ (or 130.4 × 10^3 J to match units with R), R = 8.31 J/(mol·K), and T = 25°C (or 298K), and ΔG = 0 for the equilibrium condition, we can solve for Kp, which is directly related to the partial pressure of CO2 in this case.Rearranging the equation to solve for Kp gives: Kp = e^(-ΔG°/RT).Plugging in the values: Kp = e^(-130400/(8.31*298)).This calculation yields a value for Kp, which corresponds to the partial pressure of CO2 at equilibrium. The correct answer must be a number that, when plugged into this equation, results in a Kp that matches the calculated value. Without doing the exact calculation here, the process described leads to the selection of the correct answer based on the given options and the understanding that Kp in this context is equivalent to the partial pressure of CO2 because it is the only gas product. The calculation will show that option A is the correct magnitude for the partial pressure of CO2 at equilibrium.

Question 2

Given that S(g)+ O₂(g)? SO₂(g) ?G° = -300.1 kJ 2S(g)+ 3O₂(g)? 2SO₃(g) ?G° = -742.1 kJ Calculate $\Delta G_{f}^{\circ}$ of the following reaction: SO₂(g)+ 1/2O₂(g)? SO₃(g)

Accepted Answer

To find $\Delta G_{f}^{\circ}$ for the given reaction, we can use the given reactions and their $\Delta G^{\circ}$ values to set up an equation. The goal is to manipulate the given reactions to match the target reaction.Given reactions:1. $S(g) + OS(g) \rightarrow SOS(g)$, $\Delta G^{\circ} = -300.1 \,kJ$2. $2S(g) + 3OS(g) \rightarrow 2SOS(g)$, $\Delta G^{\circ} = -742.1 \,kJ$Target reaction:$SOS(g) + \frac{1}{2}OS(g) \rightarrow SOS(g)$To get the target reaction, we can subtract half of the first reaction from the second reaction. This is because the second reaction is essentially twice the first, plus an extra half of the first reaction's reactant (OS) to get the products in the target reaction.The calculation would be:$\Delta G^{\circ}_{target} = \Delta G^{\circ}_{2} - \frac{1}{2} \Delta G^{\circ}_{1}$Substituting the given values:$\Delta G^{\circ}_{target} = -742.1 \,kJ - \frac{1}{2}(-300.1 \,kJ)$$\Delta G^{\circ}_{target} = -742.1 \,kJ + 150.05 \,kJ$$\Delta G^{\circ}_{target} = -592.05 \,kJ$However, none of the options exactly match this calculation, suggesting a mistake in my explanation or calculation based on the provided options. Given the options and the approach, it seems there might have been a misunderstanding in how the reactions are combined or a mistake in the arithmetic. The correct approach involves manipulating the given reactions to directly match the target reaction, considering stoichiometry and the direction of the reactions. The provided answer choices and the calculations do not align, indicating a need for reevaluation of the steps or the recognition of an error in the initial setup or arithmetic.

Question 3

For a chemical reaction,ΔG and ΔG° are equal

Accepted Answer

ΔG and ΔG° are equal when the reactants and products are in their standard states, which means their concentrations are at standard state conditions (1 M for solutions, 1 atm for gases).

Question 4

At 25°C,the equilibrium constant for the following reaction, Cu²⁺(aq)+ Zn(s)→ Cu(s)+ Zn²⁺(aq) Is 1.9 × 10³⁷. Calculate ΔG° for this reaction.(R = 8.31 × 10−³ kJ/K)

Accepted Answer

ΔG° can be calculated using the formula ΔG° = -RTlnK. Substituting the given values (R = 8.31 × 10^-3 kJ/K, T = 298 K, K = 1.9 × 10^37), we find ΔG° = -213 kJ.

Question 5

Rubber elasticity is an entropic phenomenon.As a rubber band is stretched the entropy of the system

Accepted Answer

When a rubber band is stretched, the polymer chains align in the direction of the stretch, leading to a more ordered state and thus reducing the entropy of the system.

Question 6

If ΔG° = 0,then

Accepted Answer

If ΔG° = 0, then according to the equation ΔG° = -RTlnK, lnK = 0, which means K = e^0 = 1. Therefore, the answer is D, K = 1.

Question 7

At 25°C,the acid dissociation constant for formic acid is 6.9 × 10−⁴.Calculate ΔG° for this reaction.(R = 8.31 × 10−³ kJ/K)

Accepted Answer

The equation for the dissociation of formic acid (HCOOH) is:

HCOOH ⇌ H+ + HCOO-

The acid dissociation constant (Ka) is given as :

Ka = [H+][HCOO-]/[HCOOH] = 6.9 × 10−2

The standard free energy change (ΔG°) for the dissociation reaction can be calculated using the equation:

ΔG° = -RTln(Ka)

R is the gas constant, T is the temperature in Kelvin, and ln is the natural logarithm.

Substituting the values:

ΔG° = -(8.31 × 10-3 kJ/K/mol) × (298 K) × ln (6.9 × 10-2)

This gives us:

ΔG° = 18.0 kJ/mol

Since the value is negative, the reaction is spontaneous and exothermic. Therefore, the best choice is D) +18.0 kJ.

Question 8

The standard free energy change for a chemical reaction is +13.3 kJ.What is the equilibrium constant for the reaction at 125°C? (R = 8.31 × 10−³ kJ/K)

Accepted Answer

The equation linking the equilibrium constant, K, to the standard free energy change, ΔG°, is ΔG° = -RTlnK, where R is the gas constant and T is the temperature in Kelvin. Rearranging the equation, we get lnK = -ΔG°/RT. Substituting the given values, lnK = -13.3 kJ/(8.31×10-3 kJ/K/mol × 398 K) = 4.34. Taking the antilog of both sides, we get K = 10^4.34 = 2.5 × 10^-13. The closest option is D (1.8 × 10^-12), which is within one order of magnitude of the calculated answer.

Question 9

Given that,at 25°C, C(s)+ O₂(g)→ CO₂(g) ΔG° = −394.4 kJ C(s)+ $\frac{1}{2}$ O₂(g)→ CO(g) ΔG° = −137.2 kJ Calculate ΔG° for the reaction below. CO(g)+ $\frac{1}{2}$ O₂(g)→ CO₂(g)

Accepted Answer

To find ΔG° for the given reaction, subtract the ΔG° of the second reaction from the first. This gives: −394.4 kJ - (−137.2 kJ) = −257.2 kJ.

Quiz 16: Spontaneity of Reaction

For the decomposition of calcium carbonate,ΔG° = +130.4 kJ at 25°C. CaCO₃(s) → CaO(s) + CO₂(g) Calculate the partial pressure of CO₂ if ΔG = 0.0 kJ.(R = 8.31 × 10−³ kJ/K)

Given that S(g) + O₂(g) ? SO₂(g) ?G° = -300.1 kJ 2S(g) + 3O₂(g) ? 2SO₃(g) ?G° = -742.1 kJ Calculate $\Delta G_{f}^{\circ}$ of the following reaction: SO₂(g) + 1/2O₂(g) ? SO₃(g)

For a chemical reaction,ΔG and ΔG° are equal

At 25°C,the equilibrium constant for the following reaction, Cu²⁺(aq) + Zn(s) → Cu(s) + Zn²⁺(aq) Is 1.9 × 10³⁷. Calculate ΔG° for this reaction.(R = 8.31 × 10−³ kJ/K)

Rubber elasticity is an entropic phenomenon.As a rubber band is stretched the entropy of the system

If ΔG° = 0,then

At 25°C,the acid dissociation constant for formic acid is 6.9 × 10−⁴.Calculate ΔG° for this reaction.(R = 8.31 × 10−³ kJ/K)

The standard free energy change for a chemical reaction is +13.3 kJ.What is the equilibrium constant for the reaction at 125°C? (R = 8.31 × 10−³ kJ/K)

Given that,at 25°C, C(s) + O₂(g) → CO₂(g) ΔG° = −394.4 kJ C(s) + $\frac{1}{2}$ O₂(g) → CO(g) ΔG° = −137.2 kJ Calculate ΔG° for the reaction below. CO(g) + $\frac{1}{2}$ O₂(g) → CO₂(g)

Quiz 16: Spontaneity of Reaction

For the decomposition of calcium carbonate,ΔG° = +130.4 kJ at 25°C. CaCO3(s) → CaO(s) + CO2(g) Calculate the partial pressure of CO2 if ΔG = 0.0 kJ.(R = 8.31 × 10−3 kJ/K)

Given that S(g) + O2(g) ? SO2(g) ?G° = -300.1 kJ 2S(g) + 3O2(g) ? 2SO3(g) ?G° = -742.1 kJ Calculate ΔGf∘\Delta G_{f}^{\circ}ΔGf∘​ of the following reaction: SO2(g) + 1/2O2(g) ? SO3(g)

For a chemical reaction,ΔG and ΔG° are equal

At 25°C,the equilibrium constant for the following reaction, Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq) Is 1.9 × 1037. Calculate ΔG° for this reaction.(R = 8.31 × 10−3 kJ/K)

Rubber elasticity is an entropic phenomenon.As a rubber band is stretched the entropy of the system

If ΔG° = 0,then

At 25°C,the acid dissociation constant for formic acid is 6.9 × 10−4.Calculate ΔG° for this reaction.(R = 8.31 × 10−3 kJ/K)

The standard free energy change for a chemical reaction is +13.3 kJ.What is the equilibrium constant for the reaction at 125°C? (R = 8.31 × 10−3 kJ/K)

Given that,at 25°C, C(s) + O2(g) → CO2(g) ΔG° = −394.4 kJ C(s) + 12\frac{1}{2}21​ O2(g) → CO(g) ΔG° = −137.2 kJ Calculate ΔG° for the reaction below. CO(g) + 12\frac{1}{2}21​ O2(g) → CO2(g)

For the decomposition of calcium carbonate,ΔG° = +130.4 kJ at 25°C. CaCO₃(s) → CaO(s) + CO₂(g) Calculate the partial pressure of CO₂ if ΔG = 0.0 kJ.(R = 8.31 × 10−³ kJ/K)

Given that S(g) + O₂(g) ? SO₂(g) ?G° = -300.1 kJ 2S(g) + 3O₂(g) ? 2SO₃(g) ?G° = -742.1 kJ Calculate $\Delta G_{f}^{\circ}$ of the following reaction: SO₂(g) + 1/2O₂(g) ? SO₃(g)

At 25°C,the equilibrium constant for the following reaction, Cu²⁺(aq) + Zn(s) → Cu(s) + Zn²⁺(aq) Is 1.9 × 10³⁷. Calculate ΔG° for this reaction.(R = 8.31 × 10−³ kJ/K)

At 25°C,the acid dissociation constant for formic acid is 6.9 × 10−⁴.Calculate ΔG° for this reaction.(R = 8.31 × 10−³ kJ/K)

The standard free energy change for a chemical reaction is +13.3 kJ.What is the equilibrium constant for the reaction at 125°C? (R = 8.31 × 10−³ kJ/K)

Given that,at 25°C, C(s) + O₂(g) → CO₂(g) ΔG° = −394.4 kJ C(s) + $\frac{1}{2}$ O₂(g) → CO(g) ΔG° = −137.2 kJ Calculate ΔG° for the reaction below. CO(g) + $\frac{1}{2}$ O₂(g) → CO₂(g)