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???
1
Expert's answer
2013-03-13T09:23:56-0400
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
Numbers and figures are an essential part of our world, necessary for almost everything we do every day. As important…
APPROVED BY CLIENTS
"assignmentexpert.com" is professional group of people in Math subjects! They did assignments in very high level of mathematical modelling in the best quality. Thanks a lot
Comments
Leave a comment