Answer to Question #25981 in Physical Chemistry for mark
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???
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
Hi Expert, so I wasn't able to try the new version because they closed of the submission portal. Oh well,
but I want to give you my sincerest thank you for helping me throughout this assignment! Honestly, you are an absolute legend! Your line by line explanation was superb and honestly helped me understand most of the concept I learnt during my semester better than my own lecturer! Thank you for consistently helping and being persistent with the versions!
You've honestly been the greatest help and I would have been so lost if it wasn't without your expertise!
So thank you once again!