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 temperature in Kelvin

The temperature \(T = 230^{\circ}C+273 = 503\ K\)

Step2: Use the ideal gas law \(PV = nRT\) to find the partial pressures

The ideal gas law is \(P=\frac{nRT}{V}\), where \(R = 0.0821\ L\cdot atm/(mol\cdot K)\), \(V = 5.86\ L\)
For \(N_2O\), \(n = 0.61\ mol\)
\(P_{N_2O}=\frac{0.61\times0.0821\times503}{5.86}\)

$$
LATEXBLOCK0
$$

For \(H_2O\), from the stoichiometry of the reaction \(NH_4NO_3(s)
ightleftharpoons N_2O(g)+2H_2O(g)\), if \(n_{N_2O}=0.61\ mol\), then \(n_{H_2O}=2\times0.61 = 1.22\ mol\)
\(P_{H_2O}=\frac{1.22\times0.0821\times503}{5.86}\)

$$
LATEXBLOCK1
$$

Step3: Write the equilibrium constant expression

For the reaction \(NH_4NO_3(s)
ightleftharpoons N_2O(g)+2H_2O(g)\), the equilibrium constant \(K = P_{N_2O}\times P_{H_2O}^2\) (solid \(NH_4NO_3\) is not included in the \(K\) - expression)
Substitute \(P_{N_2O}=4.299\ atm\) and \(P_{H_2O}=8.5975\ atm\) into the expression:

$$
LATEXBLOCK2
$$

Answer:

\(317.85\)