Question

At T(K) the $K_c$ for the reactions (I) and (II) are $2.4 \times 10^{30}$ and 1.4 respectively. What is the $K_c$ value for: $\frac{1}{2}N_2(g) + \frac{1}{2}O_2(g) + \frac{1}{2}Br_2(g) \rightleftharpoons NOBr(g)$?

• A. $\frac{1.4}{\sqrt{2.4}} \times 10^{14}$
• B. $\frac{1.4}{\sqrt{2.4}} \times 10^{-15}$
• C. $\frac{\sqrt{1.4}}{2.4} \times 10^{-15}$
• D. $\frac{\sqrt{1.4}}{2.4} \times 10^{-16}$

Hint:

When reversing a reaction, $K$ becomes $1/K$; when multiplying coefficients by $n$, $K$ becomes $K^n$.

Answer 1

The Correct Option is B

Step 1: Note the rules for combining $K_c$.
If a reaction is reversed, $K$ becomes $1/K$. If it is multiplied by a factor, $K$ is raised to that power. If reactions are added, their $K$ values multiply.

Step 2: Read the two given reactions.
Reaction (I): $2NO \rightleftharpoons N_2 + O_2$ with $K_1 = 2.4 \times 10^{30}$. Reaction (II): $NO + \tfrac{1}{2}Br_2 \rightleftharpoons NOBr$ with $K_2 = 1.4$.

Step 3: Build the target from these.
The target is $\tfrac{1}{2}N_2 + \tfrac{1}{2}O_2 + \tfrac{1}{2}Br_2 \rightleftharpoons NOBr$. We first reverse and halve reaction (I) to get $\tfrac{1}{2}N_2 + \tfrac{1}{2}O_2 \rightleftharpoons NO$, then add reaction (II).

Step 4: Find the constant for the first piece.
Reversing gives $1/K_1$, and halving raises it to the power $\tfrac{1}{2}$: \[ \left(\frac{1}{K_1}\right)^{1/2} = \frac{1}{\sqrt{2.4 \times 10^{30}}} = \frac{1}{\sqrt{2.4} \times 10^{15}}. \]

Step 5: Multiply by $K_2$.
Adding reaction (II) multiplies in $K_2 = 1.4$: \[ K = \frac{1.4}{\sqrt{2.4} \times 10^{15}} = \frac{1.4}{\sqrt{2.4}} \times 10^{-15}. \]

Step 6: State the answer.
The required equilibrium constant is as boxed.
\[ \boxed{\dfrac{1.4}{\sqrt{2.4}} \times 10^{-15}} \]