Question

A mixture containing nitrogen, hydrogen, and iodine established the following equilibrium at 400 °C:2NH3(g) + 3I2(g) ⇌ N2(g) + 6HI(g)Use the information below to calculate Kc for this reaction.N2(g)+3H2(g)⇌2NH3(g) Kc1= 0.50 at 400CH2(g)+I2(g)⇌2HI(g Kc2= 50 at 400°C

Answer 1

Answer: The value of $K_{c}$ for this reaction is 250000.

Explanation:

The given equation is as follows.

$2NH_{3}(g) + 3I_{2}(g) \rightleftharpoons N_{2}(g) + 6HI(g)$

$N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g); K_{c_{1}} = 0.50$   ... (1)

$H_{2}(g) + I_{2}(g) \rightleftharpoons 2HI(g); K_{c_{2}} = 50$  ... (2)

To balance the atoms, multiply equation (2) by 3. Hence, the equation (2) can be re-written as follows.

$3H_{2}(g) + 3I_{2}(g) \rightleftharpoons 6HI(g); K_{c_{2}} = (50)^{3}$  ... (3)

Now, subtract equation (1) from equation (3). So, the equation formed will be as follows.

$3I_{2} - N_{2} \rightleftharpoons 6HI - 2NH_{3}$

This equation can also be re-written as follows.

$3I_{2} + 2NH_{3} \rightleftharpoons N_{2} + 6HI$

This equation is similar to the equilibrium equation given to us.

Therefore, during this subtraction the equation constants get divided as follows.

$K^{'}_{c} = \frac{K_{c_{2}}}{K_{c_{1}}}\\= \frac{(50)^{3}}{0.50}\\= 250000$

Thus, we can conclude that the value of $K_{c}$ for this reaction is 250000.