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A student ran the following reaction in the laboratory at 288 K:

2NO(g) + Br2(g) 2NOBr(g)

When she introduced 0.164 moles of NO(g) and 0.131 moles of Br2(g) into a 1.00 liter container, she found the equilibrium concentration of NOBr(g) to be 0.128 M.

Calculate the equilibrium constant, Kc, she obtained for this reaction.

Sagot :

SOLUTION:

Step 1: Calculate the initial concentration of each species.

[tex]\begin{aligned} & [\text{NO}]_i = \frac{\text{0.164 mol}}{\text{1.00 L}} = 0.164 \: M \\ & [\text{Br}_2]_i = \frac{\text{0.131 mol}}{\text{1.00 L}} = 0.131 \: M \\ & [\text{NOBr}]_i = 0 \end{aligned}[/tex]

Step 2: Summarize the concentrations of each species by using ICE table.

[tex]\begin{array}{lccccc} & 2\text{NO}(g) & + & \text{Br}_2 & \rightleftharpoons & 2\text{NOBr}(g) \\ \text{Initial} \: (M): & 0.164 & & 0.131 & & 0 \\ \text{Change} \: (M): & -2(0.064) & & -0.064 & & +2(0.064) \\ \hline \text{Equilibrium} \: (M): & 0.036 & & 0.067 & & 0.128 \\ \end{array}[/tex]

Step 3: Calculate the equilibrium constant.

[tex]\begin{aligned} K_{\text{c}} & = \frac{[\text{NOBr}]^2}{[\text{NO}]^2[\text{Br}_2]} \\ & = \frac{0.128^2}{(0.036)^2(0.067)} \\ & = 188.6862 \\ & \approx \boxed{189} \end{aligned}[/tex]

Hence, the equilibrium constant obtained by the student for the reaction is 189.

[tex]\\[/tex]

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