Question

For the reaction CH4 (g) + 3 Cl2 (g) −−)−−* CHCl3 (`) + 3 HCl(g) ∆rxnG◦ m = −308.84 kJ·mol−1 and ∆rxnH◦ m = −336.59 kJ·mol−1 at 298 K. Calculate the equilibrium constant for the reaction at (a) 20◦C and (b) 500◦C.
K(20◦C) = 7.3 × 10−55; K(400◦C) = 4.1 × 10−91

K(20◦C) = 1.4 × 1054; K(400◦C) = 2.5 × 1090

K(20◦C) = 7.3 × 10−55; K(400◦C) = 1.3 × 10−18

K(20◦C) = 1.4 × 1054; K(400◦C) = 7.6 × 1017