Question

In a reversible reaction Co2+H2--->H2O+Co ,the concentration of Co2 is 0.70 Moles per dm3 and of H2 is 80.38 mole per dm3 and the concentrations of Co and H2O are each 9.46 moles per dm3.What is the value of the equilibrium constant K
A)0.168
B)0.233
C)0.628
D)1.591

Answer 1

Answer on Question #70647 – Chemistry – Other

Task:

In a reversible reaction $\mathrm{CO}_{2} + \mathrm{H}_{2} = \mathrm{H}_{2}\mathrm{O} + \mathrm{CO}$, the concentration of $\mathrm{CO}_{2}$ is 0.70 mole/dm³ and of $\mathrm{H}_{2}$ is 80.38 mole/dm³ and the concentrations of CO and $\mathrm{H}_{2}\mathrm{O}$ are each 9.46 mole/dm³. What is the value of the equilibrium constant K.

A) 0.168;

B) 0.233;

C) 0.628;

D) 1.591.

Solution:

Reaction equation:

$$CO_2 + H_2 = CO + H_2O$$

The expression for the equilibrium constant:

$$K = \frac{[H_2O]*[CO]}{[H_2]*[CO_2]};$$

Then,

$$K = \frac{[H_2O]*[CO]}{[H_2]*[CO_2]} = \frac{9.46\,\mathrm{mole/dm^3} \cdot 9.46\,\mathrm{mole/dm^3}}{80.38\,\mathrm{mole/dm^3} \cdot 0.70\,\mathrm{mole/dm^3}} = 1.5905;$$

Answer: D) 1.591.

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