Question

The reaction: ${{\text{N}}_{2}}\text{ + 3}{{\text{H}}_{2}}\text{ }\rightleftharpoons \text{ 2N}{{\text{H}}_{3}}$ takes place at $\text{450}{}^\circ \text{C}$. 1 mole of ${{\text{N}}_{2}}$ and 2 moles of ${{\text{H}}_{2}}$ are mixed in a 1-litre vessel and 1 mole of $\text{N}{{\text{H}}_{3}}$is formed at equilibrium. Then ${{\text{K}}_{\text{c}}}$ for the above reaction is:
A. 4
B. 8
C. 16
D. 32

Answer 1

The equilibrium constant is a constant which is specific for each reaction because it depends on certain conditions such as temperature. The formula of the equilibrium constant is the ratio of the concentration of the product to the concentration of the reactant.

Complete answer:
- In the given question we have to find the equilibrium constant for the synthesis of ammonia at $\text{450}{}^\circ \text{C}$.
- It is given that the moles of ammonia is 1 when one mole of nitrogen and 2 moles of hydrogen are combined.
- Firstly, we know the balance reaction will be:
${{\text{N}}_{2}}\text{ + 3}{{\text{H}}_{2}}\text{ }\rightleftharpoons \text{ 2N}{{\text{H}}_{3}}$
- At time = 0, the no. of moles of nitrogen is 1 mole and no. of moles of hydrogen is 2 moles whereas the no. of moles of ammonia will be zero because they didn't react.
- Now, at a time t let the x moles be used in the reaction. So, the no. of moles of nitrogen will be $1\,\text{- x}$ and no. of moles of hydrogen will be $\text{2 - 3x}$.
- So, the no. of moles of ammonia will be $\text{2x}$as it is given that moles of ammonia formed is 1 so, the value of x will be:
$\begin{aligned} & 2x\,\text{= 1} \\\ & \text{x = 0}\text{.5} \\\ \end{aligned}$
- Similarly, no. of moles of hydrogen and nitrogen will:
$1\text{ - x = 1 - 0}\text{.5 = 0}\text{.5}$ and $\text{2 - 3x = 2 - 3 }\times \text{ 0}\text{.5 = 0}\text{.5}$
- Now, we will write the expression for equilibrium constant i.e. the ratio of the concentration of the product to the concentration of the reactant.
${{\text{K}}_{\text{C}}}\text{ = }\dfrac{{{\text{(N}{{\text{H}}_{3}}\text{)}}^{2}}}{{{\text{(}{{\text{H}}_{2}}\text{)}}^{3}}\text{(}{{\text{N}}_{2}}\text{)}}$
${{\text{K}}_{\text{C}}}\text{ = }\dfrac{{{\text{(1)}}^{2}}}{{{\text{(0}\text{.5)}}^{3}}\text{(0}\text{.5)}}\ \text{= 16}$.

Therefore, option C is the correct answer.

Note:
Before writing the expression for equilibrium constant one must write a balanced equation. The value of equilibrium constant tells about the concentration of reactant and product in the reaction such as if the value of ${{\text{K}}_{\text{c}}}$ is more than 1000, then it means that the concentration of the product is high.