The following reaction is found to be at equilibrium at 25°C: \(2 \mathrm{SO}_{3}(g) \leftrightarrow \mathrm{O}_{2}(g)+2 \mathrm{SO}_{2}(g) \quad \Delta H=-198 \mathrm{kJ} / \mathrm{mol}\) Which of the following would cause a reduction in the value for the equilibrium constant? (A) Increasing the amount of \(\mathrm{SO}_{3}\) (B) Reducing the amount of \(\mathrm{O}_{2}\) (C) Raising the temperature (D) Lowering the temperature

Short Answer

Expert verified
Option (C) - Raising the temperature would cause a reduction in the value for the equilibrium constant.

Step by step solution

01

- Analyze Option (A)

Increasing the amount of \(SO_{3}\) would shift the reaction toward the products to regain equilibrium. However, this doesn't impact the equilibrium constant as it is only dependent on temperature, not concentrations.
02

- Analyze Option (B)

Reducing the amount of \(O_{2}\) would also shift the reaction toward the products. But again, the equilibrium constant isn't affected by concentrations, it remains the same.
03

- Analyze Option (C)

Raising the temperature will shift an exothermic reaction toward the reactants, to absorb the extra heat. This would indeed decrease the value of the equilibrium constant K.
04

- Analyze Option (D)

Lowering the temperature will shift an exothermic reaction towards the products, as it will try to generate heat. This would increase the value of the equilibrium constant K, not decrease it.

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Most popular questions from this chapter

$$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{Cl}_{2}(g) \mapsto 2 \mathrm{NOCl}(g) \Delta G^{\circ}=132.6 \mathrm{kJ} / \mathrm{mol}$$ For the equilibrium above, what would happen to the value of \(\Delta G^{\circ}\) if the concentration of \(\mathrm{N}_{2}\) were to increase and why? (A) It would increase because the reaction would become more themodynamically favored. (B) It would increase because the reaction would shift right and create more products. (C) It would decrease because there are more reactants present. (D) It would stay the same because the value of \(K_{e q}\) would not change.

A 0.1-molar solution of which of the following acids will be the best conductor of electricity? (A) \(\mathrm{H}_{2} \mathrm{CO}_{3}\) (B) \(\mathrm{H}_{2} \mathrm{S}\) (C) \(\mathrm{HF}\) (D) \(\mathrm{HNO}_{3}\)

When calcium chloride \(\left(\mathrm{CaCl}_{2}\right)\) dissolves in water, the temperature of the water increases dramatically. Compared to \(\mathrm{CaCl}_{2},\) what must be true regarding the hydration energy of \(\mathrm{CaF}_{2} ?\) (A) It would be greater because fluoride is smaller than chloride. (B) It would be the same because the charges of fluoride and chloride are identical. (C) It would be the same because hydration energy is only dependent on the IMFs present in water. (D) It would be smaller because the molar mass of \(\mathrm{CaF}_{2}\) is smaller than that of \(\mathrm{CaCl}_{2}\) .

\(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)\) Based on the information given in the table below, what is \(\Delta H^{\circ}\) for the above reaction? \(\begin{array}{cc}{\text { Bond }} & {\text { Average bond energy }(\mathrm{kJ} / \mathrm{mol})} \\ {\mathrm{H}-\mathrm{H}} & {500} \\\ {\mathrm{O}=\mathrm{O}} & {500} \\ {\mathrm{O}-\mathrm{H}} & {500}\end{array}\) (A) \(-2,000 \mathrm{kJ}\) (B) \(-500 \mathrm{kJ}\) (C) \(+1,000 \mathrm{kJ}\) (D) \(+2,000 \mathrm{kJ}\)

When calcium chloride \(\left(\mathrm{CaCl}_{2}\right)\) dissolves in water, the temperature of the water increases dramatically. Which of the following must be true regarding the enthalpy of solution? (A) The lattice energy in \(\mathrm{CaCl}_{2}\) exceeds the bond energy within the water molecules. (B) The hydration energy between the water molecules and the solute ions exceeds the lattice energy within \(\mathrm{CaCl}_{2}\) . (C) The strength of the intermolecular forces between the solute ions and the dipoles on the water molecules must exceed the hydration energy. (D) The hydration energy must exceed the strength of the intermolecular forces between the water molecules.

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