Calculate the numerical value of the equilibrium constant, Kc, for the reaction below if the equilibrium concentrations for CO, H2 , CH4 and H2O are 0.989 M, 0.993 M, 1.078 M and 0.878 M, respectively. (calculate your answer to three sig figs) CO(g) + 3 H2(g) ⇌ CH4(g) + H2O(g)

Kc = [CH4]×[H2O] / [CO]×[H2]^3

Kc = 1.078×0.878 / (0.989×0.933^3)

Kc = 0.977

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole

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