Question

A chemist wants to find Kc for the following reaction at 709 K:2NH3(g) + 3 I2 (g) ----> 6HI(g) + N2(g)

Answer 1

Answer:

422455.41

Explanation:

Corrected from source,

Given that:-

$N2(g) + 3 H2(g)\rightarrow 2NH3(g) \ Kc_1 = 0.314$

The equilibrium constant for the reverse reaction will be the reciprocal of the initial reaction.

The value of equilibrium constant the reaction

$2NH3(g)\rightarrow N2(g) + 3 H2(g)$

is:

$Kc_{1}'=\frac{1}{0.314}$

$H2(g) + I2(g)\rightarrow 2 HI(g)\ Kc_2 = 51$

If the equation is multiplied by a factor of '3', the equilibrium constant of the reverse reaction will be the cube of the equilibrium constant  of initial reaction.

The value of equilibrium constant the reaction

$3H2(g) + 3I2(g)\rightarrow 6 HI(g)$

is:

$Kc_{2}'={51}^3$

Adding both the reactions we get the final reaction. So, the equilibrium constants must be multiplied.

The value of equilibrium constant the reaction

$2NH3(g) + 3I2(g)\rightarrow 6 HI(g) + N2(g)$

is:

$Kc'=\frac{1}{0.314}\times {51}^3$  = 422455.41