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The equilibrium constant, K c, is equal to 1.4 at 1200° K for the reaction: CO2(g)...

The equilibrium constant, K c, is equal to 1.4 at 1200° K for the reaction: CO2(g) + H2(g) ⇌ CO(g) + H2O(g) If 0.35 moles of CO2 and 0.35 moles of H2 are introduced into a 1.0-L flask, what will be the concentration of CO when equilibrium is reached? The equilibrium constant, K c, is equal to 1.4 at 1200° K for the reaction: CO2(g) + H2(g) ⇌ CO(g) + H2O(g) If 0.35 moles of CO2 and 0.35 moles of H2 are introduced into a 1.0-L flask, what will be the concentration of CO when equilibrium is reached?

A.0.19 M

B.0.24 M

C.0.22 M

D. 0.16 M

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Answer #1

since volume is 1 L, concentration will be same as number of moles

CO2(g) + H2(g) ⇌ CO(g) + H2O(g)

0.35       0.35   0       0       (initial)

0.35   -x   0.35-x   x       x       (at equilibrium)

Kc = [CO][H2O]/[CO2][H2]

1.4 = x*x / (0.35-x)*(0.35-x)

1.4 = x^2 / (0.35-x)^2

sqrt(1.4) = x / (0.35-x)

x / (0.35-x) = 1.18

x = 0.413 - 1.18*x

2.18*x = 0.413

x = 0.19 M

So,

[CO] = x = 0.19 M

Answer: A

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