Question

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

$$ mathrm { n } _ { 2 } ( g ) + 2 mathrm { h } _ { 2 } mathrm { o } ( g ) ightleftharpoons 2 mathrm { h } _ { 2 } ( g ) + 2 mathrm { no } ( g ) quad k _ { p } = 1 \times 10 ^ { - 8 } $$

he fills a reaction vessel at this temperature with 8.0 atm of nitrogen gas and 11. atm of water vapor. use this data to answer the questions in the table below.

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

if you said yes, then enter the equilibrium pressure of no at right. round your answer to 1 significant digit.

Explanation:

Step1: Set up the equilibrium expression

For the reaction \(N_{2}(g)+2H_{2}O(g)
ightleftharpoons 2H_{2}(g)+2NO(g)\), the equilibrium constant \(K_{p}=\frac{P_{H_{2}}^{2}\times P_{NO}^{2}}{P_{N_{2}}\times P_{H_{2}O}^{2}}\). Let \(x\) be the change in pressure of \(N_{2}\) at equilibrium. Then \(P_{N_{2}}=(8 - x)\) atm, \(P_{H_{2}O}=(11-2x)\) atm, \(P_{H_{2}} = 2x\) atm, and \(P_{NO}=2x\) atm. Since \(K_{p}=1\times 10^{-8}\) is very small, we can assume that \(x\) is very small. So \(P_{N_{2}}\approx8\) atm and \(P_{H_{2}O}\approx11\) atm.

Step2: Substitute into the equilibrium - constant expression

\(K_{p}=\frac{(2x)^{2}\times(2x)^{2}}{8\times11^{2}}\)
\(1\times 10^{-8}=\frac{16x^{4}}{8\times121}\)
First, simplify the right - hand side: \(\frac{16x^{4}}{8\times121}=\frac{2x^{4}}{121}\)
Then solve for \(x^{4}\): \(x^{4}=\frac{1\times 10^{-8}\times121}{2}=60.5\times 10^{-8}\)
\(x^{4}=6.05\times 10^{-7}\)
Take the fourth - root of both sides: \(x=\sqrt[4]{6.05\times 10^{-7}}\)
\(x=\sqrt[4]{605\times 10^{-9}}\)
We know that \((a\times10^{n})^{\frac{1}{4}}=a^{\frac{1}{4}}\times10^{\frac{n}{4}}\). For \(a = 605\approx625=5^{4}\) and \(n=-9\)
\(x\approx5\times10^{-\frac{9}{4}}=5\times10^{-2.25}\approx5\times10^{-2}\times10^{-0.25}\approx5\times10^{-2}\times0.56\approx3\times10^{-2}\)
\(P_{NO} = 2x\)

Answer:

yes, \(6\times10^{-2}\) atm