The enthalpy and entropy change for the reaction are 30kJ mol⁻¹ and 105 JK⁻¹ mol⁻¹ respectively. The temperature at which the reaction will be in equilibrium is
(a) 273 K
(b) 450 K
(c) 300 K
(d) 285.7 K
Answers
Answered by
5
Heya Dude!
Option is (b) 450 K
Hopes help ya
Answered by
8
Answer:
285.7
Explanation:
ΔH=30 KJ/mol
ΔS=105 J/K mol
ΔS=ΔH/T
T=30×1000 J÷105
=285.7
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