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
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.
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