Question

The following reaction was carried out in a 2.75 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 9.75 mol of C, 15.7 mol of H2O, 3.60 mol of CO, and 6.00 mol of H2, what is the reaction quotient Q? Enter the reaction quotient numerically.

Answer 1

Answer:

0.1411 is the reaction quotient Q.

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power equal to stoichiometric coefficients in a balanced chemical reaction.

It is expressed as $K_{eq}$

• K is the constant of a certain reaction when it is in equilibrium,  
• Q is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

$C(s)+H_2O(g)\rightleftharpoons CO(g)+H_2(g)$

Volume of the container = 2.75 L

$Concentration =\frac{Moles}{Volume}$

Moles of carbon = 9.75 mol

Concentration of carbon =$[C]=\frac{9.75 mol}{2.75 mol}$

Moles of water= 15.7 mol

Concentration of water=$[H_2O]=\frac{15.7 mol}{2.75 mol}$

Moles of carbon monoxide = 3.60 mol

Concentration of monoxide=$[CO]=\frac{3.60 mol}{2.75 mol}$

Moles of hydrogen = 6.00 mol

Concentration of hydrogen=$[H_2]=\frac{6.00 mol}{2.75 mol}$

The expression reaction quotient, Q is given as;

$Q=\frac{[H_2][CO]}{[C][H_2O]}$

$Q=\frac{\frac{6.00 mol}{2.75 mol}\times \frac{3.60 mol}{2.75 mol}}{\frac{9.75 mol}{2.75 mol}\times \frac{15.7 mol}{2.75 mol}}$

$Q=0.1411$

0.1411 is the reaction quotient Q.