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

Question

Elisha Jacobs

Chemistry

10 November, 02:29

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

+3

Answers (1)

Know the Answer?

Sparks · Answer 1

(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)

For the following reaction2 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. 0.521b. 0.494c. 2.03d. 1.57

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