Question

The standard enthalpy of certain reaction is approximately constant at +125kJmol-1 from 800K to 1500K. The standard Gibbs energy is +25kJmol-1 at 1150K. Estimate the temperature at witch the equilibrium constant becomes greater than 1.

Answer 1

Answer:

T = 3006.976 K

Explanation:

∴ ΔH° = 125 KJ/mol ( 800K - 1500K)

∴ ΔG° = 25 KJ/mol (1150 K)

⇒ T = ? ∴ K > 1

In the equilibrium:

• K = e∧(ΔG°/RT)

∴ K > 1

If ΔG°/RT = 1

⇒ e∧(1) = 2.72 > 1

∴ ΔG°/RT = 1

⇒ T = ΔG°/R = (25 KJ/mol)/(8.314 E-3 KJ/mol.K)

⇒ T = 3006.976 K

verifying:

• K = e∧(ΔG°/RT)

⇒ K = e∧(25/((8.314 E-3)(3006.976)))

⇒ K = 1.000000061 > 1