Question

Calculate the equilibrium constant at 298 K for the reaction of formaldehyde (CH2O) with hydrogen gas using the following information. CH2O(g) + 2H2(g) LaTeX: \longleftrightarrow⟷ CH4(g) + H2O(g) LaTeX: \DeltaΔH°= –94.9 kJ; LaTeX: \DeltaΔS°= –224.2 J/K A. 1.92 B. 9.17 x 10-6 C. 2.07 x 1028 D. 1.10 x 105 E. 8.08 x 104 F. 3.98 x 1011 Group of answer choices

Answer 1

Answer:

E. 8.08 x 10⁴.

Explanation:

Hello,

In this case, for the reaction:

$CH_2O(g) + 2H_2(g) \rightleftharpoons CH_4(g) + H_2O(g)$

We can compute the Gibbs free energy of reaction via:

$\Delta G\°=\Delta H\°-T\Delta S\°$

Since both the entropy and enthalpy of reaction are given at 298 K (standard temperature), therefore:

$\Delta G\°=-94.9kJ-(298K)(-224.2\frac{J}{K}*\frac{1kJ}{1000kJ} )\\\\\Delta G\°=-28.1kJ$

Then, as the equilibrium constant is computed as:

$K=exp(-\frac{\Delta G\°}{RT} )$

We obtain:

$K=exp(-\frac{-28.1kJ/mol}{8.314x10^{-3}\frac{kJ}{mol* K}}*298K )\\\\K=8.08 x10^4$

For which the answer is E. 8.08 x 10⁴.

Best regards,