Calculate value of -ln-K-eq- the reaction 250 K-N-2O-4 -g- rightleftharpoons 2NO-2 -g-Given- H-0-f-NO-2-g- - - 40-407 kJ - molH-0-f-N-2O-4-g- - - 70 kJ - molS-0-r - 10 JK-1-4-41-2-1-2

Question

Toppr Admin · Accepted Answer

-G0-H0-x2212-TS0-H0r-H0f-p-x2212-H0f-R-2-xD7-40-407-x2212-70-10-814kJ-G0-10-814kJ-x2212-250-xD7-10J-K-10-814kJ-x2212-2500J-8314J-G0-x2212-RTlnK8314-x2212-8-314-xD7-250lnKlnK-x2212-4-

Calculate value of $^{'} l n (K_{e q})$ ' for the reaction at 250 K.
$N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$
Given: $H_{f}^{0} ((N O_{2}) g) = + 40.407 k J / m o l$
$H_{f}^{0} ((N_{2} O_{4}) g) = + 70 k J / m o l$
$S_{r}^{0} = 10 J K^{- 1}$

4
-4
1.2
-1.2

$+ G^{0} = + H^{0} - T S^{0}$
$+ H_{r}^{0} = (+ H_{f}^{0}) p - (+ H_{f}^{0})_{R}$
$= 2 \times (+ 40.407) - (+ 70)$
$= + 10.814 k J$
$+ G^{0} = + 10.814 k J - 250 \times 10 J / K$
$= + 10.814 k J - 2500 J = 8314 J$
$+ G^{0} = - R T l n K$
$8314 = - 8.314 \times 250 l n K$
$l n K = - 4.$

Calculate value of ′ln(Keq)' for the reaction at 250 K.N2O4(g)⇌2NO2(g)Given: H0f((NO2)g)=+40.407kJ/molH0f((N2O4)g)=+70kJ/molS0r=10JK−14-41.2-1.2

+G0=+H0−TS0+H0r=(+H0f)p−(+H0f)R=2×(+40.407)−(+70)=+10.814kJ+G0=+10.814kJ−250×10J/K=+10.814kJ−2500J=8314J+G0=−RTlnK8314=−8.314×250lnKlnK=−4.

Calculate value of $^{'} l n (K_{e q})$ ' for the reaction at 250 K.
$N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$
Given: $H_{f}^{0} ((N O_{2}) g) = + 40.407 k J / m o l$
$H_{f}^{0} ((N_{2} O_{4}) g) = + 70 k J / m o l$
$S_{r}^{0} = 10 J K^{- 1}$

4
-4
1.2
-1.2

$+ G^{0} = + H^{0} - T S^{0}$
$+ H_{r}^{0} = (+ H_{f}^{0}) p - (+ H_{f}^{0})_{R}$
$= 2 \times (+ 40.407) - (+ 70)$
$= + 10.814 k J$
$+ G^{0} = + 10.814 k J - 250 \times 10 J / K$
$= + 10.814 k J - 2500 J = 8314 J$
$+ G^{0} = - R T l n K$
$8314 = - 8.314 \times 250 l n K$
$l n K = - 4.$