Question

1. The following reaction was carried out in a 3.75 L reaction vessel at 1100 K:

C(s)+H2O(g)⇌CO(g)+H2(g)

If during the course of the reaction, the vessel is found to contain 9.25 mol of C, 14.8 mol of H2O, 4.00 mol of CO, and 7.60 mol of H2, what is the reaction quotient Q?

2. The reaction 2CH4(g)⇌C2H2(g)+3H2(g) has an equilibrium constant of K = 0.154.

If 6.00 mol of CH4, 4.65 mol of C2H2, and 11.20 mol of H2 are added to a reaction vessel with a volume of 5.10 L , what net reaction will occur? (Left, right, or no futher reaction)

Answer 1

1. [H2] = 7.6 / 3.750 = 2.026 M
[CO] = 4.00 / 3.750 = 1.066 M
[H2O] = 14.8 / 3.750 = 3.946M
Q = [CO] [H2] / [H2O] = 1.066x2.026/3.946 =0.5473
The concentration of C in the solid state is a constant

2. 2CH4(g)⇌C2H2(g)+3H2(g)

given data,

equilibrium constant of K = 0.154. ,6.00 mol of CH4, 4.65 mol of C2H2, and 11.20 mol of H2

2CH4(g)⇌ C2H2(g)+ 3H2(g)

Initial : 6 moles 0 0

at eqbr : 6-x x 3x

but the x value is x= 4.65 moles

in products side more nuber of moles are formed , so that the concentration or moles of reacttants are very less. so that to attain equilibrium the products again converted into reactants , by forming backward reaction Rb. so that the net reaction will occur towards the left