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Question: For reversible reaction:, \(X_{(g)} + 3Y_{(g)} \rightleftharpoons 2Z_{(g);}\Delta H = - 40kJ\) Sta...

For reversible reaction:,

X(g)+3Y(g)⇌2Z(g);ΔH=−40kJX_{(g)} + 3Y_{(g)} \rightleftharpoons 2Z_{(g);}\Delta H = - 40kJ

Standard entropies of X, Y and Z are 60, 40 and 50kJmol−150kJmol^{- 1} respectively. The temperature at which the above reaction is in equilibrium is

A

273 K

B

600 K

C

500 K

D

400 K

Answer

500 K

Explanation

Solution

: X + 3Y 2Z

ΔS=2×50−(60+3×40)=−80kJ\Delta S = 2 \times 50 - (60 + 3 \times 40) = - 80kJ

ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S when ΔG=0\Delta G = 0

T=ΔHΔS=−40×1000−80=500KT = \frac{\Delta H}{\Delta S} = - 40 \times \frac{1000}{- 80} = 500K