**The value of log K for the reaction A⇋ B at 298 K is ……. (Nearest integer)Given: ∆H ^{0} = – 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:

Solution:

∆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
^{₀}= -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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