Question #25981

2A(g)+B(g)⇌2C(g)At equilibrium a 6 L container contains 6.0 mol of A, 0.18 mol of B and 3.6 mol of C gases. At a constant temperature the volume of the container is decreased to 4 L and some C gas is added. When the new equilibrium is established, the mole number of B is found to be 0.3. What is the mole number of C gas added into the container???

Expert's answer

Equilibrium constant is:

K = [C]^2/[A]^2*[B] = (3.6/6)^2 / (6.0/6)^2 * (0.18/6) = 0.36/0.03 = 12

w = 4/6 = 0.667

New K is 12*0.667 = 8

6.0/6-x 0.3 3.6/6+x

2A(g) + B(g) ⇌ 2C(g)

x = 0.37 M

real number of moles is 0.37*4 = 1.48 mol

K = [C]^2/[A]^2*[B] = (3.6/6)^2 / (6.0/6)^2 * (0.18/6) = 0.36/0.03 = 12

w = 4/6 = 0.667

New K is 12*0.667 = 8

6.0/6-x 0.3 3.6/6+x

2A(g) + B(g) ⇌ 2C(g)

x = 0.37 M

real number of moles is 0.37*4 = 1.48 mol

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