Question

1. nh₃ is normally encountered as a gas with a pungent odor. it is formed from hydrogen and nitrogen (equation given below). a chemist pours a mixture of nitrogen gas and hydrogen gas in a 1.0 litre reaction vessel in 1:3 ratio. the equilibrium constant for the reaction is 1.0 × 10⁻⁴, and the equilibrium value of h₂(g) is 0.12 mol/l. calculate the equilibrium value of nh₃.

n₂(g) + 3 h₂(g) ⇔ 2 nh₃(g)

Explanation:

Step1: Recall the equilibrium constant expression

For the reaction \(\ce{N_{2}(g) + 3H_{2}(g) ightleftharpoons 2NH_{3}(g)}\), the equilibrium constant \(K_{c}\) is given by the formula:
\[K_{c}=\frac{[\ce{NH_{3}}]^{2}}{[\ce{N_{2}}][\ce{H_{2}}]^{3}}\]
We know \(K_{c} = 1.0\times10^{-4}\), \([\ce{H_{2}}]=0.12\space mol/L\), and we need to find \([\ce{NH_{3}}]\). First, we need to find \([\ce{N_{2}}]\) using the initial ratio and the change in concentration.

Step2: Determine the initial concentration of \(\ce{N_{2}}\) and \(\ce{H_{2}}\)

The initial ratio of \(\ce{N_{2}}\) to \(\ce{H_{2}}\) is \(1:3\). Let the initial concentration of \(\ce{N_{2}}\) be \(x\) and \(\ce{H_{2}}\) be \(3x\). At equilibrium, the concentration of \(\ce{H_{2}}\) is \(0.12\space mol/L\). Let the change in concentration of \(\ce{N_{2}}\) be \(y\), then the change in concentration of \(\ce{H_{2}}\) is \(3y\) (from the stoichiometry of the reaction: 1 mole of \(\ce{N_{2}}\) reacts with 3 moles of \(\ce{H_{2}}\)). So, the equilibrium concentration of \(\ce{H_{2}}\) is \(3x - 3y=0.12\), and the equilibrium concentration of \(\ce{N_{2}}\) is \(x - y\). Notice that \(3x - 3y = 3(x - y)\), so \(x - y=\frac{0.12}{3}=0.04\space mol/L\). So, \([\ce{N_{2}}]=0.04\space mol/L\) at equilibrium.

Step3: Substitute values into the \(K_{c}\) expression

We have \(K_{c}=\frac{[\ce{NH_{3}}]^{2}}{[\ce{N_{2}}][\ce{H_{2}}]^{3}}\), substitute \(K_{c} = 1.0\times10^{-4}\), \([\ce{N_{2}}]=0.04\space mol/L\), and \([\ce{H_{2}}]=0.12\space mol/L\) into the formula:
\[1.0\times10^{-4}=\frac{[\ce{NH_{3}}]^{2}}{(0.04)(0.12)^{3}}\]
First, calculate the denominator: \((0.04)(0.12)^{3}=0.04\times0.001728 = 6.912\times10^{-5}\)

Step4: Solve for \([\ce{NH_{3}}]^{2}\)

Multiply both sides of the equation by \(6.912\times10^{-5}\):
\[[\ce{NH_{3}}]^{2}=1.0\times10^{-4}\times6.912\times10^{-5}=6.912\times10^{-9}\]

Step5: Solve for \([\ce{NH_{3}}]\)

Take the square root of both sides:
\[[\ce{NH_{3}}]=\sqrt{6.912\times10^{-9}}=\sqrt{69.12\times10^{-10}}=\sqrt{69.12}\times10^{-5}\approx8.31\times10^{-5}\space mol/L\] Wait, no, let's do the calculation again. Wait, the denominator calculation: \((0.12)^{3}=0.12\times0.12\times0.12 = 0.001728\), then \(0.04\times0.001728 = 0.00006912=6.912\times10^{-5}\). Then \(K_{c}\times\) denominator \(=1.0\times10^{-4}\times6.912\times10^{-5}=6.912\times10^{-9}\). Then \([\ce{NH_{3}}]=\sqrt{6.912\times10^{-9}}=\sqrt{69.12\times10^{-10}}=\sqrt{69.12}\times10^{-5}\approx8.31\times10^{-5}\)? Wait, no, \(\sqrt{6.912\times10^{-9}}=\sqrt{6.912}\times\sqrt{10^{-9}}\approx2.63\times3.162\times10^{-5}\)? Wait, no, \(10^{-9}=10^{-4.5\times2}=(10^{-4.5})^{2}\), but better to write \(6.912\times10^{-9}=69.12\times10^{-10}\), so \(\sqrt{69.12\times10^{-10}}=\sqrt{69.12}\times10^{-5}\approx8.31\times10^{-5}\)? Wait, no, \(\sqrt{69.12}\approx8.31\), so \([\ce{NH_{3}}]\approx\sqrt{1.0\times10^{-4}\times0.04\times(0.12)^{3}}\). Let's compute inside the square root: \(1.0\times10^{-4}\times0.04 = 4\times10^{-6}\), \((0.12)^{3}=0.001728\), then \(4\times10^{-6}\times0.001728 = 6.912\times10^{-9}\), then square root of \(6.912\times10^{-9}\) is \(\sqrt{6.912}\times\sqrt{10^{-9}}\approx2.63\times3.162\times10^{-5}\)? Wait, no, \(\sqrt{10^{-9}} = 10^{-4.5}=10^{-4}\times10^{-0.5}\approx10^{-4}\times0.3162 = 3.162\times10^{-5}\), then \(2.63\times3.162\times10^{-5}\approx8.31\times10^{-5}\space mol/L\). Wait, but let's check the steps again.

Wait, actually, the initial ratio is \(1:3\), so at equilibrium, the ratio of \([\ce{N_{2}}]\) to \([\ce{H_{2}}]\) should still be \(1:3\) (because the stoichiometry is 1:3, so the change in concentration is also in 1:3 ratio). So, if \([\ce{H_{2}}]=0.12\), then \([\ce{N_{2}}]=\frac{0.12}{3}=0.04\space mol/L\) (since the ratio of \(\ce{N_{2}}\) to \(\ce{H_{2}}\) at equilibrium is 1:3, same as initial, because the reaction consumes them in 1:3 ratio). That's a simpler way: because the reaction is \(\ce{N_{2} + 3H_{2} ightleftharpoons 2NH_{3}}\), the ratio of \(\ce{N_{2}}\) to \(\ce{H_{2}}\) at any time (initial or equilibrium) should be \(1:3\) if we start with only \(\ce{N_{2}}\) and \(\ce{H_{2}}\) in 1:3 ratio. So, \([\ce{N_{2}}]=\frac{[\ce{H_{2}}]}{3}=\frac{0.12}{3}=0.04\space mol/L\).

Now, substitute into \(K_{c}\) expression:
\[1.0\times10^{-4}=\frac{[\ce{NH_{3}}]^{2}}{(0.04)(0.12)^{3}}\]
Solve for \([\ce{NH_{3}}]^{2}\):
\[[\ce{NH_{3}}]^{2}=1.0\times10^{-4}\times0.04\times(0.12)^{3}\]
Calculate \((0.12)^{3}=0.12\times0.12\times0.12 = 0.001728\)
Then \(0.04\times0.001728 = 0.00006912 = 6.912\times10^{-5}\)
Then \(1.0\times10^{-4}\times6.912\times10^{-5}=6.912\times10^{-9}\)
Then \([\ce{NH_{3}}]=\sqrt{6.912\times10^{-9}}=\sqrt{69.12\times10^{-10}}=\sqrt{69.12}\times10^{-5}\approx8.31\times10^{-5}\space mol/L\) (Wait, no, \(\sqrt{6.912\times10^{-9}}=\sqrt{6.912}\times\sqrt{10^{-9}}\approx2.63\times3.162\times10^{-5}\)? Wait, no, \(10^{-9}=10^{-4.5\times2}=(10^{-4.5})^{2}\), but \(10^{-4.5}=10^{-4}\times10^{-0.5}\approx10^{-4}\times0.3162 = 3.162\times10^{-5}\), and \(\sqrt{6.912}\approx2.63\), so \(2.63\times3.162\times10^{-5}\approx8.31\times10^{-5}\space mol/L\). Wait, but let's check with exact calculation:

\(6.912\times10^{-9}=6.912e - 9\)

\(\sqrt{6.912e - 9}=\sqrt{6.912}\times\sqrt{1e - 9}\approx2.63\times3.16227766e - 5\approx8.31e - 5\space mol/L\)

Wait, but let's do the calculation again:

\([\ce{NH_{3}}]^{2}=1.0\times10^{-4}\times0.04\times(0.12)^{3}\)

\((0.12)^{3}=0.001728\)

\(0.04\times0.001728 = 0.00006912\)

\(1.0\times10^{-4}\times0.00006912 = 6.912\times10^{-9}\)

\([\ce{NH_{3}}]=\sqrt{6.912\times10^{-9}}\approx8.31\times10^{-5}\space mol/L\) (approximately \(8.3\times10^{-5}\space mol/L\) or more accurately, let's compute \(\sqrt{6.912\times10^{-9}}\):

\(6.912\times10^{-9}=69.12\times10^{-10}\)

\(\sqrt{69.12}\approx8.31\)

\(\sqrt{10^{-10}} = 10^{-5}\)

So \([\ce{NH_{3}}]\approx8.31\times10^{-5}\space mol/L\approx8.3\times10^{-5}\space mol/L\)

Answer:

The equilibrium value of \([\ce{NH_{3}}]\) is approximately \(\boldsymbol{8.3\times10^{-5}\space mol/L}\) (or more precisely \(8.31\times10^{-5}\space mol/L\))