Question

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

Kc = __________

Answer 1

422455.41

Step-by-step 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)'=(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'=(1)/(0.314)* {51}^3` = 422455.41