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Find the Kc value for the reaction at 12C when Kp=0.082 3A(s) + 2B(g) + C...

Find the Kc value for the reaction at 12C when Kp=0.082

3A(s) + 2B(g) + C (g) -> 2 D (g)
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Answer #1

3A(s) + 2B(g) + C (g) -> 2 D (g)

Kp = Kc(RT)n

n = no of moles of products - no of moles of reactants in gases

       = 2 -(2+1) = -1

T = 12C0 = 12+273 = 285K

R = 0.0821L-atm/mole-k

Kp = 0.082

Kp = Kc(RT)n

0.082 = Kc(0.0821*285)-1

Kc = 1.92 >>>> answer

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