Question

Calculate the value of the equilibrium constant, Kc, for the reaction Q(g) + X(g) <------> 2M(g) + N(g) given that

M(g) <------> Z(g)      Kc1=3.98

6R(g) <-------> 2N(g) + 4Z(g)    Kc2=0.442

3X(g) + 3Q(g) <--------> 9R(g)   Kc3=13.6

Answer 1

M(g) <------> Z(g)      Kc1=3.98

Equilibrium constant , Kc1 = [Z]/[M] --------------(1)

6R(g) <-------> 2N(g) + 4Z(g)    Kc2=0.442

Equilibrium constant , Kc2 = ([N]²[Z]⁴) / [R]⁶ ---------------(2)

3X(g) + 3Q(g) <--------> 9R(g)   Kc3=13.6

Equilibrium constant , Kc 3 = [R]⁹ / ([X]3 [Q]³) -------------(3)

Q(g) + X(g) <------> 2M(g) + N(g)

Equilibrium constant , Kc = ([M]²[N])/([Q][X])

= {1 /([Z]/[M])}2 x { ([N]²[Z]⁴) / [R]⁶ }^1/2 x {[R]⁹ / ([X]3 [Q]³)}^1/3

= (1/(Kc1)²) x (Kc2)^1/2 x (Kc3)^1/3

= (1/3.98)² x (0.442)^1/2 x (13.6)^1/3

= 0.1002