Question

the equilibrium constant, ( k_{c} ), for the following reaction is ( 7.00 \times 10^{-5} ) at ( 673 mathrm{~k} ). calculate ( k_{p} ) for this reaction at this temperature.

mathrm{nh}_{4} mathrm{i}(s)
ightarrow mathrm{nh}_{3}(g)+mathrm{hi}(g)

( k_{mathrm{p}}= )

Explanation:

Step1: Identify the formula relating \(K_p\) and \(K_c\)

The formula is \(K_p = K_c(RT)^{\Delta n}\), where \(R = 0.0821\ L\cdot atm/(mol\cdot K)\), \(T\) is the temperature in Kelvin, and \(\Delta n\) is the change in the number of moles of gas.

Step2: Calculate \(\Delta n\)

For the reaction \(NH_4I(s)\to NH_3(g)+HI(g)\), the number of moles of gas on the reactant side (\(n_{reactant}\)) is \(0\) (since \(NH_4I\) is solid), and on the product side (\(n_{product}\)) is \(2\) (\(1\) mole of \(NH_3\) and \(1\) mole of \(HI\)). So, \(\Delta n=n_{product}-n_{reactant}=2 - 0=2\).

Step3: Substitute the values into the formula

Given \(K_c = 7.00\times10^{-5}\), \(T = 673\ K\), \(R=0.0821\ L\cdot atm/(mol\cdot K)\), and \(\Delta n = 2\).

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Answer:

\(0.214\)