Consider the following gas-phase dissociation reaction: X (g) Y (g) + Z (g) Given U^ = 83.35 kJ mol ⁻¹ and S^ = 175 J K ⁻¹ mol ⁻¹ . Assuming U^ and S^ are independent of temperature, calculate the temperature above which the reaction becomes spontaneous. (Given : R = 8.3 J K ⁻¹ mol ⁻¹ )

Step-by-step solution

For a reaction to be spontaneous, the standard Gibbs free energy change must be negative ( G^ G^ = H^ - T S^ The standard enthalpy change H^ is related to U^ by: H^ = U^ + n_g R T Substituting this into the Gibbs equation gives: U^ + n_g R T - T S^ Rearranging for T : U^ T > U^ S^ - n_g R For the given reaction X (g) Y (g) + Z (g) , the change in gaseous moles is: n_g = (1 + 1) - 1 = 1 Now, calculate the denominator term (ensure units are consistent, converting J to kJ): S^ - n_g R = 175 - (1 8.3) = 166.7 J K ⁻¹ mo