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Question No. 6 Bookmark QuestionFor the reaction,A(g) + B(g) → C(g) + D(g), ΔH0 and ΔS0 are, respectively, –29.8 kJ mol–1 and –0.100 kJ K–1 mol–1 at 298K. The equilibrium constant for the reaction at 298K is

Question

Question No. 6 Bookmark QuestionFor the reaction,A(g) + B(g) → C(g) + D(g), ΔH0 and ΔS0 are, respectively, –29.8 kJ mol–1 and –0.100 kJ K–1 mol–1 at 298K. The equilibrium constant for the reaction at 298K is

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Solution

To solve this problem, we need to use the equation for Gibbs Free Energy (ΔG) and the relationship between ΔG and the equilibrium constant (K).

Step 1: Calculate ΔG using the equation ΔG = ΔH - TΔS.

Given: ΔH = -29.8 kJ/mol = -29800 J/mol (since 1 kJ = 1000 J) ΔS = -0.100 kJ/K/mol = -100 J/K/mol T = 298 K

Substitute these values into the equation:

ΔG = -29800 J/mol - (298 K * -100 J/K/mol) = -29800 J/mol + 29800 J/mol = 0 J/mol

Step 2: Use the relationship between ΔG and K, which is ΔG = -RTlnK, where R is the gas constant (8.314 J/K/mol) and T is the temperature in Kelvin.

Since ΔG = 0, we can rearrange the equation to solve for K:

0 = -RTlnK 0 = -8.314 J/K/mol * 298 K * lnK 0 = lnK

To get rid of the natural logarithm (ln), we take the exponential of both sides:

e^0 = K 1 = K

So, the equilibrium constant for the reaction at 298K is 1.

This problem has been solved

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