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At \(450 \mathrm{~K}, \mathrm{~K}_{\mathrm{p}}=2.0 \times 10^{10}\) bar for the given reaction at equilibrium.
What is \(K_c\) at this temperature ?
What is \(K_c\) at this temperature ?
Solution:
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Verified Answer
For the given reaction, \(\Delta \mathrm{n}_{\mathrm{g}}=2-3=-1\)
\(\begin{aligned}
&K_p=K_c(R T)^{A n} \text { or } K_c=K_p(R T)^{-\Delta n}=K_p(R T) \\
&=\left(2.0 \times 10^{10} \text { bar }^{-1}\right)\left(0.0831 \mathrm{~L} \text { bar K }{ }^{-1} \mathrm{~mol}^{-1}\right)(450 \mathrm{~K}) \\
&=74.8 \times 10^{10} \mathrm{~L} \mathrm{~mol}^{-1}=7.48 \times 10^{11} \mathrm{~L} \mathrm{~mol}^{-1}
\end{aligned}\)
\(\begin{aligned}
&K_p=K_c(R T)^{A n} \text { or } K_c=K_p(R T)^{-\Delta n}=K_p(R T) \\
&=\left(2.0 \times 10^{10} \text { bar }^{-1}\right)\left(0.0831 \mathrm{~L} \text { bar K }{ }^{-1} \mathrm{~mol}^{-1}\right)(450 \mathrm{~K}) \\
&=74.8 \times 10^{10} \mathrm{~L} \mathrm{~mol}^{-1}=7.48 \times 10^{11} \mathrm{~L} \mathrm{~mol}^{-1}
\end{aligned}\)
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