Question

the equilibrium constant, ( k_p ), for the following reaction is ( 1.04\times10^{-2} ) at 548 k.
( nh_4cl(s)
ightleftharpoons nh_3(g)+hcl(g) )
if an equilibrium mixture of the three compounds in a 4.37 l container at 548 k contains 3.08 mol of ( nh_4cl(s) ) and 0.117 mol of ( nh_3(g) ), the partial pressure of ( hcl(g) ) is (square) atm.

Explanation:

Step1: Write the expression for \(K_p\)

For the reaction \(NH_4Cl(s)
ightleftharpoons NH_3(g)+HCl(g)\), \(K_p = P_{NH_3}\times P_{HCl}\) (since the partial pressure of solids is not included in \(K_p\) expression).

Step2: Calculate the partial pressure of \(NH_3\)

Use the ideal gas law \(PV = nRT\). Rearranged for \(P\), \(P=\frac{nRT}{V}\).
Given \(n = 0.117\space mol\), \(R=0.0821\space L\cdot atm/(mol\cdot K)\), \(T = 548\space K\), \(V=4.37\space L\)
\(P_{NH_3}=\frac{0.117\times0.0821\times548}{4.37}\)

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Step3: Solve for \(P_{HCl}\)

Since \(K_p = 1.04\times 10^{-2}\) and \(K_p=P_{NH_3}\times P_{HCl}\)
\(P_{HCl}=\frac{K_p}{P_{NH_3}}\)
Substitute \(K_p = 1.04\times 10^{-2}\) and \(P_{NH_3}\approx1.205\space atm\)
\(P_{HCl}=\frac{1.04\times 10^{-2}}{1.205}\)

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Answer:

\(8.63\times 10^{-3}\)