The value of log K for the reaction A⇋ B at 298 K is ……. (Nearest integer)
Given: ∆H0 = – 54.07 kJ mol–1
∆S° = 10 JK–1 mol–1
(Take 2.303 × 8.314 × 298 = 5705)


Answer: 10
Solution:
The question is asking for the value of log K for the reaction A ⇋ B at 298 K, rounded to the nearest integer. In chemical equilibrium, the equilibrium constant (K) is related to the change in Gibbs free energy (∆G) through the equation:

∆G = ∆H – T∆S

Where:

  • ∆G is the change in Gibbs free energy.
  • ∆H is the change in enthalpy.
  • T is the temperature in Kelvin.
  • ∆S is the change in entropy.

Given data:

  • ∆H = -54.07 kJ/mol
  • ∆S° = 10 J/K/mol
  • T = 298 K
  • The conversion factor for units: 2.303 × 8.314 × 298 = 5705

Now, let’s calculate the value of log K using the given data and the equation:

∆G = ∆H – T∆S

∆G = (-54070 J/mol) – (298 K) × (10 J/K/mol)

∆G = (-54070 J/mol) – (2980 J/mol)

∆G = -57050 J/mol

Next, we can relate ∆G to the equilibrium constant K using the equation:

∆G = -2.303RT × log K

Now, plug in the values:

  • ∆G = -57050 J/mol
  • R is the gas constant = 8.314 J/(mol·K)
  • T = 298 K

So, let’s rearrange the equation:

  • (-57050 J/mol) = (-2.303 × 8.314 J/(mol·K) × 298 K) × log K

Now, solve for log K:

log K = (-57050 J/mol) / (-2.303 × 8.314 J/(mol·K) × 298 K)

log K ≈ 10 So, the value of log K for the reaction A ⇋ B at 298 K is approximately 10 when rounded to the nearest integer.

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