Question

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the equilibrium constant, ( k_{p} ), for the following reaction is 0.497 at 500 k.
calculate ( k_{c} ) for this reaction at this temperature.
( mathrm{pcl}_{5}(g)
ightarrow mathrm{pcl}_{3}(g)+mathrm{cl}_{2}(g) )
( k_{c}= )

Explanation:

Step1: Determine $\Delta n$

For the reaction \(PCl_{5}(g)\to PCl_{3}(g)+Cl_{2}(g)\), \(\Delta n=(1 + 1)-1=1\) (moles of gaseous products - moles of gaseous reactants).

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

Rearrange for \(K_{c}\): \(K_{c}=\frac{K_{p}}{(RT)^{\Delta n}}\). Given \(K_{p} = 0.497\), \(R=0.0821\space L\cdot atm/(mol\cdot K)\), \(T = 500\space K\), and \(\Delta n=1\).
Substitute values: \(K_{c}=\frac{0.497}{(0.0821\times500)^{1}}\)
Calculate denominator: \(0.0821\times500 = 41.05\)
Then \(K_{c}=\frac{0.497}{41.05}\approx0.0121\)

Answer:

\(0.0121\)