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.
1 answer:
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 > 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∧(25/((8.314 E-3)(3006.976)))
⇒ K = 1.000000061 > 1
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