Question

The following reaction was carried out in a 3.00 L reaction vessel at 1100 K:

C(s)+H2O(g) --> CO(g)+H2(g)

If during the course of the reaction, the vessel is found to contain 5.25 mol of C, 12.2 mol of H2O, 3.90 mol of CO, and 7.90 mol of H2, what is the reaction quotient Q?

Answer 1

Answer:

$Q=0.840$

Explanation:

Hello,

In this case, since the given undergoing chemical reaction is correctly balanced, the reaction quotient is computed as well as the equilibrium constant but in terms of the given concentrations that are:

$C_{C}=\frac{5.25mol}{3.00L} =1.75M\\C_{H_2O}=\frac{12.2mol}{3.00L} =4.07M\\C_{CO}=\frac{3.90mol}{3.00L}=1.30M\\C_{H_2}=\frac{7.90mol}{3.00L} =2.63M$

In such a way, the reaction quotient turns out:

$Q=\frac{C_{CO}C_{H_2}}{C_{H_2O}}=\frac{1.30M*2.63M}{4.07M}\\ Q=0.840$

Taking into account that carbon is not included since it is solid.

Best regards.