# Question 06971

Nov 29, 2015

$\text{0.0390 M}$

#### Explanation:

So, you know that your equilibrium reaction has an equilibrium constant equal to $2.3 \cdot {10}^{- 4}$.

Right from the start, you can use the value of the reaction's equilibrium constant to predict that the equilibrium concentration of hydrogen cyanide, $\text{HCN}$, will be very small compared with thos of nitrogen gas ans acetylene.

As you know, when ${K}_{e q} < 1$, the reaction favors the reactants at equilibrium, which means that you can expect most of the reactants to remain unreacted once equilibrium is set.

Set up an ICE table to help you calculate the equilibrium concentration of hydrogen cyanide

${\text{ " "N"_text(2(g]) " "+" " "C"_2"H"_text(2(g]) " "rightleftharpoons" " color(red)(2)"HCN}}_{\textrm{\left(g\right]}}$

color(purple)("I") " " " "3.3" " " " " " " "2.0" " " " " " " " " " " "0
color(purple)("C") " "(-x)" " " " " "(-x)" " " " " " " "(+color(red)(2)x)
color(purple)("E") " "3.3-x" " " " " "2.0-x" " " " " " " "color(red)(2)x

By definition, the equilibrium constant for this reaction will be equal to

${K}_{e q} = \left(\left[{\text{HCN"]^color(red)(2))/( ["N"_2] * ["C"_2"H}}_{2}\right]\right)$

In your case, this will be equal to

${K}_{e q} = {\left(\textcolor{red}{2} x\right)}^{\textcolor{red}{2}} / \left(\left(3.3 - x\right) \left(2.0 - x\right)\right) = \frac{4 {x}^{2}}{\left(3.3 - x\right) \left(2.0 - x\right)}$

Now, because ${K}_{e q}$ is so small, you can approximate

$3.3 - x \approx 3.3 \text{ }$ and $\text{ } 2.0 - x \approx 2.0$

This means that you have

$2.3 \cdot {10}^{- 4} = \frac{4 {x}^{2}}{3.3 \cdot 2.0}$

${x}^{2} = \frac{2.3 \cdot 3.3 \cdot 2.0}{4} \cdot {10}^{- 4}$

$x = \sqrt{\frac{2.3 \cdot 3.3 \cdot 2.0}{4} \cdot {10}^{- 4}} = 0.01948$

This means that the equilibrium concentrations of the chemical species that take part in the reaction will be

["N"_2] = 3.3 - 0.01948 = "3.28 M"

["C"_2"H"_2] = 2.0 - 0.01948 = "1.98 M"#

$\left[\text{HCN"] = color(red)(2) * 0.01948 = color(green)("0.0390 M}\right)$

I'll leave the answers rounded to three sig figs.

The initial prediction turned out to be very accurate - the equilibrium concentration of the product is much smaller than the equilibrium concentrations of the two reactants.