For Cl₂O(g)+ 3/2 O₂(g)→ 2 ClO₂ △_rH° = 126 kJ/mol,and Δ_rS° = -74.9 J/(mol K)at 377 °C.What is K_eq? A)0.97 B)6.12 × 10^-7 C)4.27 × 10^-22 D)9.17 × 10^-15 E)1.07 × 10¹⁴

Question

Quizplus · Accepted Answer

Using the equation ΔG1U1B1r= ΔH1U1B1r−TΔS1U1B1r, where T is the temperature in Kelvin, we can calculate ΔG1U1B1r as:
ΔG1U1B1r= (126,000 J/mol) - (377 K)(-74.9 J/(mol K))
ΔG1U1B1r= 151,923 J/mol

Then, we can use the equation ΔG1U1B1r= -RTlnK to solve for K:
-151,923 J/mol = -(8.314 J/(mol K))(377 K)lnK
lnK = -21.51
K = e^(-21.51) = 9.17 × 10^-15

Therefore, the answer is D, 9.17 × 10^-15.

For Cl2O(g)+ 3/2 O2(g)→ 2 ClO2 △rH° = 126 KJ/mol,and

For Cl₂O(g)+ 3/2 O₂(g)→ 2 ClO₂ △_rH° = 126 KJ/mol,and