Question

At ${445}^{\circ }C,K_c$ for the following reaction is 0.020.
$2HI\left(g\right)\rightleftharpoons H_2\left(g\right)+I_2\left(g\right)$
A mixture of $H_2,I_2$, and HI in a vessel at ${445}^{\circ }C$ has the following concentrations: $\left[HI\right]=2.0M,\left[H_2\right]=0.50Mand\left[I_2\right]=0.10M$. Which one of the following statements concerning the reaction quotient, $Q_c$, is TRUE for the above system?

• A. $Q_c=K_c$ the system is at equilibrium.
• B. $Q_c$ is less than $K_c$; more $H_2$ and $I_2$ will be produced.
• C. $Q_c$ is less than $K_c$; more HI will be produced.
• D. $Q_c$ is greater than $K_c$; more $H_2$ and $I_2$ will be produced.

Answer 1

The correct option is B $Q_c$ is less than $K_c$; more $H_2$ and $I_2$ will be produced.

For the given reaction,
$2HI\left(g\right)\rightleftharpoons H_2\left(g\right)+I_2\left(g\right)$
The reaction guetient $Q_c$ can be calculate as:
$Q_c=\frac{H_2\left[I_2\right]}{[HI{]}^2}$
substituting the given values, we get
$Q_C=\frac{\left[0.50\right]\left[0.10\right]}{[2.0{]}^2}$
$Q_c=\frac{0.05}{4.0}$
$Q_c=0.0125$
Given $K_c=0.020$ So, $Q_c<K_c$
For the condition $Q_c<K_c.$
The concentration of the reactants is higher than it would be at equilibrium and the concentration of product is too low. So, in order to reach the equilibrium, the reaction will favor the forward direction meaning that more $H_2$ and $I_2$ will be produced