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  • At 450K, K_p= 2.0 × 10^10/bar for the given reaction at equilibrium. 2 SO_2 (g) + O_2 (g) 2 SO_3 (g) What is K_c at this temperature ?

7.10     At 450K, Kp= 2.0 × 1010 /bar for the given reaction at equilibrium.

                2SO_{2}_{(g)}+O_{2}_{(g)}\rightleftharpoons 2SO_{3}_{(g)} 

               What is Kc at this temperature?

Answers (1)

best_answer

We have,
K_p= 2\times 10^{10}/bar
We know that the relation between K_p and K_c;

K_p=K_c(RT)^{\Delta n} 
Here \Delta n = ( moles of product) - (moles of  reactants)

2SO_{2}_{(g)}+O_{2}_{(g)}\rightleftharpoons 2SO_{3}_{(g)}

So. here \Delta n = 2-3 = -1

By applying the formula we get;

2\times 10^{10} =K_c(0.0831 L\ bar/K/mol)\times 450K
K_c= \frac{2\times 10^{10}}{(0.0831 L\ bar/K/mol)\times 450K}

        =74.79 \times 10^{10}L mol^{-1}
         =  7.48\times 10^{10}L mol^{-1}

Posted by

manish

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