Answer to Question #105278 in General Chemistry for Hanna

Question

For the reaction below, the enthalpy of reaction is ΔrH = 188.6 kJ mol−1 and the thermodynamic equilibrium constant is K = 3.21×10−19 at 298.15 K. What is the value of K at 613 K?

NH4Br(s) ⟶ NH3(g) + HBr(g)

Answer 1

$\Delta G^0=-RTlnK$ , at K=3.21*10^-19, $\Delta$ G=105504. Jmole-1

$\Delta G^0=\Delta H^0 - T\Delta S$ , −TΔS = -83095,6

ΔS=-278,7 Jmole-1K-1

K at T=613Kalv

Δ G=17754Jmole-1

K_equive=32.63

We assume that the enthalpy and entropy are temperature independent