Q.

Standard entropy of X2 ,Y2and XY3 are 60, 40 and 50 JK-1mol-1 respectively, for this reaction 12X2 +32Y2⇌XY3,∆H=-30KJto be at equilibrium,the temperature should be

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a

300 K

b

750 K

c

1000 K

d

1250 K

answer is B.

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Detailed Solution

12X2 +32Y2⇌XY3∆S0=∑Sproducts0-∑Sreactents0=50-40×32 + 12×60=50-60+30=-40KJmole-1 ∆G0=∆H0-T∆S0 , at equilibrium ∆G0 =0 ∆H0=T∆S0 ; T =∆H0∆S0=-30×103-40=750K
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