Question

If the equilibrium constant for the decomposition of acetylene (C2H2) to solid carbon and molecular hydrogen exeeds 10, less than 80% of the reactant will be converted to products. True or False?

Answer 1

C2H2 (g) ---> 2C(s) + 2H2 (g) is teh equation

Let initially we have 1 mole C2H2

then at equilibirum C2H2 moles = 1-X , H2 moles = 2X

Keq= [H2]^2 /[C2H2]    ( solids not involved in K expression)

10 = ( 2X)^2 / ( 1-X)          ( since they mention K exceeds 10 , we take Keq =10)

4X^2 +10X-10 = 0

X= 0.7655

% of C2H2 converted = 100 ( C2H2 dissociated) / initial C2H2 = 100 x0.7655 /1 = 76.55 %

if we take Keq = 20 which is above 10

we get X % = 85 %

Thus overall statement is false since when Keq is 20 which is above 10 , we get 85 % convertion.