CH4 (g) + H2O (g) <- -> CO (g) + 3 H2 (g)             Kc=5.2 at 1768 K

a) What is the equilibrium expression for this reaction in terms of Kc?
b) Suppose that the initial concentrations of all species at 1768 K are:
[CH4] = 0.210 M, [H2O] = 0.171 M, [CO] = 0.533 M, and [H2] = 0.987 M
         I. Determine the value for the reaction quotient.
         II. In what direction will the reaction proceed at this temperature? Justify your answer.
c) Calculate the value of the equilibrium constant, Kp, for this reaction at 1768 K.
d) Calculate the value for the free energy change, Delta G^O 1768 K, of this reaction. Include units with your answer.
e) The equilibrium concentrations of CO and H2 increase when the temperature of the system increases. Is the forward reaction exothermic or endothermic? Justify your answer.
f) A chemist who works for a company that produces hydrogen gas suggests that they could increase the yield of H2 (g) by adjusting the volume of the equilibrium system. Should the volume be increased or decreased? Explain.
g) Calculate the value of Kc for the following reaction.
CO (g) + 3 H2 (g) <- -> CH4 (g) + H2O (g)

1) kc = [CH4][H2O]/[CO][H2]^3
2)Calculate Qc using the same expression
check if Qc>Kc or less than Kc
if it is greater it means more products are there so it will shift to left and if it is less it means more reactants are there so it will shift to the right .
3)kp = kc *RT^(delta n)
here delta n = 4-2=2
4)delta G =-RTln K
5)If the reaction is going in the direction of the products on increasing temperature it means it is endothermic in nature.why?Try to figure out using le chatelier's principle
6.volume is inversely proportional to pressure.so if we increase volume it will shift in the direction of less pressure ie less number of moles(reactants).Figure out what will happen if you decrease volume?
7.New Kc =1/ 5.2 because equation is reversed.
6.