Question

${{K}_{C}}$ for the reaction: $\dfrac{1}{2}{{N}_{2}}_{(g)}+\dfrac{1}{2}{{O}_{2(g)}}+\dfrac{1}{2}B{{r}_{2}}(g)\rightleftharpoons NOB{{r}_{(g)}}$ is $9\times {{10}^{-4x}}$. Find x (at $298{}^\circ K$).
Given that
$\begin{align}
& {{K}_{C}}=2.4\times {{10}^{30}}\text{ for 2N}{{\text{O}}_{(g)}}\rightleftharpoons {{N}_{2}}_{(g)}+{{O}_{2(g)}} \\
& {{K}_{C}}=1.4\text{ for N}{{\text{O}}_{(g)}}+\dfrac{1}{2}B{{r}_{2(g)}}\rightleftharpoons NOB{{r}_{(g)}} \\
\end{align}$

Answer 1

Hint: ${{K}_{c}}$ denotes the equilibrium constant of the chemical equation which gives the relationship between the products and reactants when a chemical reaction reaches equilibrium. It is basically the ratio of the concentration of products to the concentration of the reactants, each raised to the power of their respective stoichiometric coefficients.

Complete step by step solution:
-Here we are given two equations along with their rate constant,
$\begin{align}
& {{K}_{{{C}_{1}}}}=2.4\times {{10}^{30}}\text{ for 2N}{{\text{O}}_{(g)}}\rightleftharpoons {{N}_{2}}_{(g)}+{{O}_{2(g)}}...(1) \\
& {{K}_{{{C}_{2}}}}=1.4\text{ for N}{{\text{O}}_{(g)}}+\dfrac{1}{2}B{{r}_{2(g)}}\rightleftharpoons NOB{{r}_{(g)}}...(2) \\
\end{align}$
-We will rearrange the above two equations in such a way that we will get the equation whose rate
constant is to be calculated, that is-
$\dfrac{1}{2}{{N}_{2}}_{(g)}+\dfrac{1}{2}{{O}_{2(g)}}+\dfrac{1}{2}B{{r}_{2}}(g)\rightleftharpoons NOB{{r}_{(g)}}...(i)$
The product of the second equation is the same as of the equation (i), so we need not reverse the equation (2). The coefficient of $B{{r}_{2}}$ in equation (2) is also the same as that of equation (i), so we need not multiply it with any number.
But, the NO in reaction (2) is an extra term which can be removed by reversing the equation (1) followed by the multiplication of equation (1) and (2), we will get a new equation which is equation (3),
$\begin{align}
& {{N}_{2}}+{{O}_{2}}\to 2NO...(3) \\
& {{K}_{{{c}_{3}}}}=\dfrac{1}{{{K}_{{{c}_{1}}}}}=\dfrac{1}{1.24\times {{10}^{30}}} \\
\end{align}$

Now, the NO from the reactants is removed but it is now present in the product side of equation (3). To remove NO, we need to multiply the equation (3) by $\dfrac{1}{2}$, which will give us a new equation, equation (4).
$\begin{align}
& \dfrac{1}{2}{{N}_{2}}+\dfrac{1}{2}{{O}_{2}}\to NO...(4) \\
& {{K}_{{{c}_{4}}}}=\sqrt{{{K}_{{{c}_{3}}}}}=\sqrt{\dfrac{1}{2.4\times {{10}^{30}}}}=0.6455\times {{10}^{-15}} \\
\end{align}$
-Now, we have got our desired equation as equation (4). Equation (4) and equation (2) on multiplication gives us equation (5), which is
$\dfrac{1}{2}{{N}_{2}}_{(g)}+\dfrac{1}{2}{{O}_{2(g)}}+\dfrac{1}{2}B{{r}_{2}}(g)\rightleftharpoons NOB{{r}_{(g)}}...(5)$
This equation is the same as equation (i), which is the equation of which rate constant is to be calculated.
The rate constant of this equation is calculated as,
${{K}_{c}}={{K}_{{{c}_{4}}}}\times {{k}_{{{c}_{2}}}}=(0.644\times {{10}^{-15}})\times 1.4=9.037\times {{10}^{-16}}$
-We need to give the value of x, in $9\times {{10}^{-4x}}$, which is can now be calculated from the ${{K}_{c}}$ value calculated above.

Therefore, the value of x is 4.

Note: There are characteristics of equilibrium constant which are given below-
(i) Equilibrium constant for a specific reaction at a constant temperature is fixed.
(ii) A catalyst changes only the rate of forwarding and backward reactions. It does not affect the value of the equilibrium constant.
(iii) Equilibrium may be affected by changes in concentration, pressure, temperature, inert gases favouring either forward or backward reaction but not the equilibrium constant.
(iv) In the case of stepwise multiple equilibria leading to the final products, the equilibrium constant for the net equilibrium is equal to the product of each stepwise equilibrium constant.