JEE MainChemistryThermodynamics (C)
For a specific chemical reaction carried out at 27^ C , the standard free energy change ( G^ ) is -2.5 kJ mol ⁻¹ and the standard entropy change ( S^ ) is 15 J K ⁻¹ mol ⁻¹ . If the standard enthalpy change ( H^ ) for the reaction is x 10^2 J mol ⁻¹ , the value of x is ______.
Correct answer
20
Step-by-step solution
Given: Temperature, T = 27^ C = 27 + 273 = 300 K G^ = -2.5 kJ mol ⁻¹ = -2500 J mol ⁻¹ S^ = 15 J K ⁻¹ mol ⁻¹ The standard Gibbs free energy equation is: G^ = H^ - T S^ Rearranging for standard enthalpy change ( H^ ): H^ = G^ + T S^ Substitute the given values: H^ = -2500 + (300 15) H^ = -2500 + 4500 H^ = 2000 J mol ⁻¹ Comparing this with x 10^2 J mol ⁻¹ : 2000 = 20 10^2 Therefore, x = 20 . Answer: 20