Question

Question 6.17 For the reaction at 298 K,

2A + B→C

ΔH = 400 kJ mol–1 and ΔS = 0.2 kJ K–1 mol–1

At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature range?

Class XI Thermodynamics Page 183

Answer 1

given, ΔH = 400 KJ/mol, ΔS = 0.2 KJ/K/mol 
        we know, the formula of Gibbs free energy,
  ΔG = ΔH - TΔS 
        = 400KJ/mol - T × 0.2 KJ/mol
now, equate ΔG = 0 
     400 = T × O.2 
   ⇒T = 2000K
we know, for spontanteous reaction, ΔG < 0 
               Hence, T > 2000 K, the reaction will be spontaneous.

Answer 2

Hi

Here is your answer,

Given, ΔH = 400 k .mol ⁻¹ , ΔS = 0.2 kJ⁻¹

Gibbs free energy , ΔG = ΔH - TΔS

                  0 = 400 kJ mol⁻¹ - T × 0.2 kJ K⁻¹ mol⁻¹

Temperature, T = 400 kJ mol⁻¹/0.2 kJ K⁻¹ mol⁻¹    


                                 = 2000 K