The relation between Kp and Kc is given as:
\( K_{p}=K_{c}(RT)^{\Delta n}\)
(i) Given,
R = 0.0831 bar mol–1K–1
\( \Delta n = 3-2 =1\)
T = 500 K
Kp=\( 1.8\times 10^{-2}\)
Now,
Kp = Kc (RT) ∆n
\( \Rightarrow 1.8\times 10^{-2}=K_{c}(0.0831\times 500)^{1}\\ \\ \)
\( \Rightarrow K_{c}=\dfrac{1.8\times 10^{-2}}{0.0831\times 500}\\ \\\)
\( =4.33\times 10^{-4}(approximately)\)
(ii) Here,
∆n =2 – 1 = 1
R = 0.0831 bar mol–1K–1
T = 1073 K
Kp= 167
Now,
Kp = Kc (RT) ∆n
\( \Rightarrow 167=K_{c}{(0.0831\times 1073)^{\Delta n}}\\ \\ \Rightarrow K_{c}=\dfrac{167}{0.0831\times 1073}\\ \\ =1.87(approximately)\)
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