2SO_2 (g) + O_2 (g) doubleheadarrow 2SO_3 (g) T = 500 degree C K_p = 48.2 (c)H_2

Question

2SO_2 (g) + O_2 (g) doubleheadarrow 2SO_3 (g) T = 500 degree C K_p = 48.2 (c)H_2O (t) doubleheadarrow H_2O (g) T = 60 degree C K_p = 0.196 atm (d) CoO (s) + CO (g) doubleheadarow Co (s) + CO_2 (g) T = 550 degree C K_p = 4.90 times 10^2 (e) CH_3 NH_2 (aq) + H_2 O (l).rightarrow CH_3 NH_3 + (aq) + OH^-(aq) T = 25 degree C K_p = 8.7times 10^-9 Calculate the equilibrium constant at 25 degree C for each of the following reactions from the value of delta G degree given. (a) O_2 (g) + 2F_2 (g) rightarrow 2 OF_2 (g) delta G degree = - 9.2 KJ (b) I_2 (s) + Br_2 (l) rightarrow 2IBr (g) delta G degree = 7.3 KJ (c) 2LiOH (s) + CO_2 (g) rightarrow Li_2 CO_3 (s) + H_2 O (g) delta G degree = -79 KJ (d) N_2 O_3 (g) rightarrow NO (g) + NO_2 (g) delta G degree = -1.6 KJ (e) S_nCl_4 (l) rightarrow delta G degree = 8.0 KJ Calculate the equilibrium constant at 25 degree C for each of the following reactions from the value of delta G degree given (a).I_2 (s) + Cl_2 (g) rightarrow 2ICI (g

Explanation / Answer

Question 41

G = -RT ln Keq

R = gas constant = 8.314 J K-1 Mole -1

T = temperature in Kelvin

.Part a

-9.2 x 103 = -8.314 x 298 ln Keq

Keq = 40.96

Part B

7.3 x 103 = -8.314 x 298 ln Keq

Keq =0.005255

Part C

-79 x 103 = -8.314 x 298 ln Keq

Keq = 7 x 1013

Part D

-1.6 x 103 = -8.314 x 298 ln Keq

Keq = 1.907

Part E

-8 x 103 = -8.314 x 298 ln Keq

Keq = 25.23

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