Answer:
a) 48KJ
b) -48KJ
Explanation:
Given that;
ln(K2/K1) = ΔH°/R(1/T2 - 1/T1)
K2= equilibrium constant at T2
K1 = equilibrum constant at T1
R = gas constant
T1 = initial temperature
T2 = final temperature
When we double the equilibrium constant K1; K2 = 2K1
T1 = 310 K
T2 = 310 + 15 = 325 K
ln(2K1/K1) =- ΔH°/R(1/T2 - 1/T1)
ln2 = -ΔH°/8.314(1/325 - 1/310)
0.693 = -ΔH°/8.314(3.08 * 10^-3 - 3.2 * 10^-3)
0.693 = -ΔH°/8.314 (-0.00012)
0.693 = 0.00012ΔH°/8.314
0.693 * 8.314 = 0.00012ΔH°
ΔH° = 0.693 * 8.314/0.00012
ΔH° = 48KJ
b) K2 =K1/2
ln(K1/2/K1) =- ΔH°/R(1/T2 - 1/T1)
ln (1/2) = -ΔH°/8.314 (1/325 - 1/310)
-0.693 = -ΔH°/8.314 (-0.00012)
-0.693 = 0.00012ΔH°/8.314
-0.693 * 8.314 = 0.00012ΔH°
ΔH°= -0.693 * 8.314/0.00012
ΔH°= -48KJ
