Question

use the references to access important values if needed for this
the equilibrium constant, ( k_{c} ), for the following reaction is ( 5.10 \times 10^{-6} ) at ( 548 mathrm{~k} ).
calculate ( k_{mathrm{p}} ) for this reaction at this temperature.
mathrm{nh}_{4} mathrm{cl}(s)
ightarrow mathrm{nh}_{3}(g)+mathrm{hcl}(g)
( k_{mathrm{p}}= )

Explanation:

Step1: Determine the value of $\Delta n$

For the reaction \(NH_{4}Cl(s)
ightarrow NH_{3}(g)+HCl(g)\), the number of moles of gaseous products minus the number of moles of gaseous reactants. Reactants (\(NH_{4}Cl\)) is solid (\(n = 0\) for solid in terms of gas - phase mole change). So \(\Delta n=(1 + 1)-0=2\)

Step2: Use the formula \(K_{p}=K_{c}(RT)^{\Delta n}\)

The ideal gas constant \(R = 0.0821\space L\cdot atm/(mol\cdot K)\), \(T = 548\space K\), \(K_{c}=5.10\times 10^{-6}\) and \(\Delta n = 2\)
Substitute the values into the formula:

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First calculate \(0.0821\times548 = 44.9908\)
Then \((44.9908)^{2}=44.9908\times44.9908\approx2024.17\)

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Answer:

\(K_{p}=0.0103\)