chemistry

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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].

  • chemistry -

    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*RTdelta n

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