Question

Before the reaction occurs, the concentration of A is 0.071M. If the concentration of A at equilibrium is 0.03195M, what is the equilibrium constant? A(g) = 2B(g) + C(g)​

Answer 1

Answer:

Assume that the initial concentration of both B and C are zero.

Kc = 7.5 × 10⁻³.

Explanation:

What's the expression for the equilibrium constant of this reversible reaction? Note that the steps here apply only to equilibriums where all species are either gaseous (g) or in an aqueous solution (aq). Solids and liquids (water in particular) barely influence the equilibrium; in many cases they should not appear in the expression for the equilibrium constant.

Raise the concentration of each product to their coefficient's power.

• The coefficient in front of B is 2. Raise the concentration of B to the second power, which will give $[\text{B}]^{2}$.
• The coefficient in front of C is 1. Raise the concentration of C to the first power, which is the same as $[\text{C}]$.

The product of these terms will be the numerator in the expression of the equilibrium constant. For this reaction, the numerator shall be $[\text{B}]^{2} \cdot[\text{C}]$.

Repeat these steps for the reactants:

• The coefficient in front of A is 1. Raise the concentration of A to the first power, which is the same as $[\text{A}]$.

If there are more than one reactants, multiple each of those terms. The product shall be the denominator of the expression for the equilibrium constant. For this reaction, the denominator shall be $[\text{A}]$.

Hence the equilibrium constant:

$\displaystyle K_c = \frac{[\text{B}\;(g)]^{2} \cdot[\text{C}\;(g)]}{[\text{A}\;(g)]}$,

where $[\text{A}]$, $[\text{B}]$, and $[\text{C}]$ are concentrations when the reaction is at equilibrium.

Construct a RICE table to find the equilibrium concentration of each species. R stands for reaction, I for initial conditions, C for change in concentration, and E for equilibrium conditions. Assume that only A is initially present in the system. Let the decrease in the concentration of $\text{A}$ be $x\;\text{M}$.

$\begin{array}{c|ccccc}\text{R}&\text{A}\;(g) &\rightleftharpoons &2\;\text{B}\;(g)& +& \text{C}\;(g)\\ \text{I}& 0.071 \;\text{M} & & 0\;\text{M} && 0 \;\text{M} \\ \text{C} &-x \;\text{M}&&+2x\;\text{M} & & +x\;\text{M}\\\text{E}&0.071 - x\;\text{M} && 2x\;\text{M} & &x\;\text{M}  \end{array}$.

The coefficient in front of B is twice that in front of A. In other words, for each unit of A consumed, two units of B are produced. Assume that volume stays the same. The decrease in the concentration of $\text{A}\;(g)$ is $x\;\text{M}$. Accordingly, the increase in the concentration of $\text{B}\;(g)$ shall be $2x\;\text{M}$.

The question states that the equilibrium concentration of $A\;(g)$ is $0.03195\;\text{M}$.

$0.071 - x = 0.03195$,

$x = 0.03905$.

At equilibrium:

• $[\text{A}\;(g)] = 0.03195\;\text{M}$ according to the question;
• $[\text{B}\;(g)] = 2 x \;\text{M}= 0.0781\;\text{M}$ as in the RICE table;
• $[\text{C}\;(g)] = x = 0.03905\;\text{M}$ as in the RICE table.

Hence the equilibrium constant.

$\displaystyle \begin{aligned}K_c &= \frac{[\text{B}\;(g)]^{2} \cdot[\text{C}\;(g)]}{[\text{A}\;(g)]}\\&=\frac{{0.0781}^{2} \times 0.03905}{0.03195}\\&=7.5\times 10^{3}\;\text{M}\end{aligned}$.