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For the following reaction
2 R (g) --> Q (g)
if the equilibrium concentration of R is 0.776 M and that of Q is 1.22 M at 25 ?C, what is the value of the equilibrium constant at this temperature?

a. 0.521
b. 0.494
c. 2.03
d. 1.57

Respuesta :

princessesther2011

Answer:

(C) 2.03

Explanation:

Kc (equilibrium constant) = [Q]/[R]^2

[Q] = 1.22 M

[R] = 0.776 M

Kc = 1.22/0.776^2 = 1.22/0.602176 = 2.03 (to 2 decimal places)

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