Question

c. A student put 1.18 mole of substance A and and 2.85 mole of substance B into a 10 litre flask which was then closed. The reaction that took place was: A(g)+ 2B(g) 3C(g) + D(g)
On analysis the equilibrium mixture at 25 0 C was found to contain 0.376mole of D. Calculate Kc and Kp at this temperature.

[ R = 0.0821atm l mole-1K-1].

Answer 1

These are in 1 L; therefore, mols = M.
........A + 2B ==> 3C + D
I .....1.18..2.85...0....0
C......-x....-2x...3x....x
E...1.18-x..2.85-x..3x...x
x in the problem = 0.376 for D at equilibrium.
Calculate concns of each of the others and substitute into Keq expression and solve for Kc.
Then convert top Kp = Kc*RT^{delta n}

Answer 2

I need help

Answer 3

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