Question

a chemist is studying the following equilibrium, which has the given equilibrium constant at a certain temperature:

2ch₄(g) ⇌ c₂h₂(g) + 3h₂(g) kₚ = 3.×10⁻⁸

he fills a reaction vessel at this temperature with 13. atm of methane gas. use this data to answer the questions in the table below.

can you predict the equilibrium pressure of c₂h₂, using only the tools available to you within aleks? yes no

if you said yes, then enter the equilibrium pressure of c₂h₂ at right. round your answer to 1 significant digit. atm

Explanation:

Step1: Set up the ICE table

Let \(x\) be the change in pressure of \(C_{2}H_{2}\).

	\(2CH_{4}(g)\)	\(C_{2}H_{2}(g)\)	\(3H_{2}(g)\)
Change	\(-2x\)	\(+x\)	\(+3x\)
Equilibrium	\(13 - 2x\)	\(x\)	\(3x\)

Step2: Write the expression for \(K_{p}\)

\(K_{p}=\frac{P_{C_{2}H_{2}}\times P_{H_{2}}^{3}}{P_{CH_{4}}^{2}}\)
Substitute the equilibrium pressures: \(3\times10^{-8}=\frac{x\times(3x)^{3}}{(13 - 2x)^{2}}\)

Since \(K_{p}\) is very small (\(3\times 10^{-8}\)), \(13-2x\approx13\)
So the equation becomes \(3\times10^{-8}=\frac{x\times27x^{3}}{169}\)
\(3\times10^{-8}=\frac{27x^{4}}{169}\)
\(x^{4}=\frac{3\times10^{-8}\times169}{27}\)
\(x^{4}\approx1.878\times10^{-7}\)
\(x=\sqrt[4]{1.878\times 10^{-7}}\)
\(x\approx0.017\)

Answer:

yes, \(0.02\) atm