The value of log K for the reaction A⇌B at 298 K is _______. (Nearest integer)Given: ΔH°=−54.07 kJ mol−1ΔS°=10JK−1 mol−1(Taken 2.303×8.314×298=5705 )

To solve this problem, we need to use the formula for the Gibbs free energy change (ΔG°) and the relationship between ΔG° and the equilibrium constant (K).

Step 1: Calculate ΔG° using the formula ΔG° = ΔH° - TΔS°

Given: ΔH° = -54.07 kJ mol-1 = -54070 J mol-1 (since 1 kJ = 1000 J) ΔS° = 10 J K-1 mol-1 T = 298 K

Substitute these values into the formula:

ΔG° = -54070 J mol-1 - (298 K * 10 J K-1 mol-1) ΔG° = -54070 J mol-1 - 2980 J mol-1 ΔG° = -57050 J mol-1

Step 2: Use the relationship between ΔG° and K, which is ΔG° = -RT ln K, where R is the gas constant and T is the temperature in Kelvin. We can rearrange this formula to solve for K: K = e^(-ΔG°/RT)

Given: R = 8.314 J K-1 mol-1 (gas constant) T = 298 K ΔG° = -57050 J mol-1

Substitute these values into the formula:

K = e^(-(-57050 J mol-1) / (8.314 J K-1 mol-1 * 298 K)) K = e^(57050 J mol-1 / 2494.572 J mol-1) K = e^22.86

Step 3: To find log K, we can use the natural logarithm of K (ln K) and convert it to log K using the relationship log K = ln K / ln 10.

log K = 22.86 / ln 10 log K = 22.86 / 2.303 log K = 9.92

Rounding to the nearest integer, the value of log K for the reaction A⇌B at 298 K is 10.