Question

101.7 g of solid nh₄no₃ is heated in a 5.86 l container at 230.°c and the following reaction occurs: nh₄no₃(s) ⇌ n₂o(g) + 2h₂o(g) at equilibrium there are 0.61 moles of n₂o present. what is the value of the equilibrium constant, k, for this reaction? enter your answer to 4 decimal places.

Explanation:

Step1: Calculate the moles of \(NH_4NO_3\)

The molar mass of \(NH_4NO_3\) is \(M=(14 + 4\times1+14 + 3\times16)\space g/mol=80\space g/mol\).
The moles of \(NH_4NO_3\), \(n=\frac{m}{M}=\frac{101.7\space g}{80\space g/mol}=1.27125\space mol\)

Step2: Determine the moles of \(H_2O\) at equilibrium

From the balanced equation \(NH_4NO_3(s)
ightleftharpoons N_2O(g)+2H_2O(g)\), if \(n(N_2O) = 0.61\space mol\), then \(n(H_2O)=2\times n(N_2O)=2\times0.61 = 1.22\space mol\)

Step3: Calculate the concentrations

The temperature \(T=(230 + 273.15)\space K=503.15\space K\), and \(V = 5.86\space L\)
Using the ideal - gas law \(PV=nRT\), or \(P=\frac{nRT}{V}\) (where \(R = 0.0821\space L\cdot atm/(mol\cdot K)\))
For \(N_2O\): \(P_{N_2O}=\frac{n_{N_2O}RT}{V}=\frac{0.61\space mol\times0.0821\space L\cdot atm/(mol\cdot K)\times503.15\space K}{5.86\space L}\)

$$P_{N_2O}=\frac{0.61\times0.0821\times503.15}{5.86}\space atm\approx4.29\space atm$$

For \(H_2O\): \(P_{H_2O}=\frac{n_{H_2O}RT}{V}=\frac{1.22\space mol\times0.0821\space L\cdot atm/(mol\cdot K)\times503.15\space K}{5.86\space L}\)

$$P_{H_2O}=\frac{1.22\times0.0821\times503.15}{5.86}\space atm\approx8.58\space atm$$

Step4: Calculate the equilibrium constant \(K_p\)

The equilibrium constant expression for \(NH_4NO_3(s)
ightleftharpoons N_2O(g)+2H_2O(g)\) is \(K_p = P_{N_2O}\times(P_{H_2O})^2\)

$$K_p=4.29\times(8.58)^2$$

$$K_p=4.29\times73.6164$$

$$K_p = 316.8143$$

Answer:

\(316.8143\)