Question

Consider the following equilibrium, $$ 2 \text{NO}(g) \rightleftharpoons \text{N}_2 (g) + \text{O}_2 (g); K_1 = 2.4 \times 10^{20} $$ $$ \text{NO}(g) + \frac{1}{2} \text{Br}_2 (g) \rightleftharpoons \text{NOBr}(g); K_2 = 1.4 $$ Calculate $ K_C $ for the reaction, $$ \frac{1}{2} \text{N}_2 (g) + \frac{1}{2} \text{O}_2 (g) + \frac{1}{2} \text{Br}_2 (g) \rightleftharpoons \text{NOBr}(g) $$

• A. $9.0 \times 10^{-16}$
• B. \( 9.48 \times 10^{-9} \)
• C. \( 8.08 \times 10^{-12} \)
• D. \( 8.96 \times 10^{11} \)

Hint:

When manipulating equilibrium reactions, remember to adjust the equilibrium constant based on the stoichiometric changes (e.g., doubling a reaction squares the constant).

Answer 1

The Correct Option is A

To determine the equilibrium constant (K_C) for the reaction between nitrogen, oxygen, and bromine to form NOBr, we'll analyze the given equilibrium systems step-by-step.

1. Given Equilibrium Reactions:
(a) $$ 2\text{NO}(g) \rightleftharpoons \text{N}_2(g) + \text{O}_2(g) $$
with equilibrium constant:
$$ K_{c_1} = \frac{[\text{N}_2][\text{O}_2]}{[\text{NO}]^2} = 2.4 \times 10^{30} $$
(b) $$ \text{NO}(g) + \frac{1}{2}\text{Br}_2(g) \rightleftharpoons \text{NOBr}(g) $$
with equilibrium constant:
$$ K_{c_2} = \frac{[\text{NOBr}]}{[\text{NO}][\text{Br}_2]^{1/2}} = 1.4 $$

2. Target Reaction:
We want to find K_C for:
$$ \frac{1}{2}\text{N}_2(g) + \frac{1}{2}\text{O}_2(g) + \frac{1}{2}\text{Br}_2(g) \rightleftharpoons \text{NOBr}(g) $$
with equilibrium expression:
$$ K_C = \frac{[\text{NOBr}]}{[\text{N}_2]^{1/2}[\text{O}_2]^{1/2}[\text{Br}_2]^{1/2}} $$

3. Relationship Between Constants:
The target reaction can be obtained by:
1) Taking the reverse of reaction (a) and halving it
2) Adding reaction (b)
Therefore:
$$ K_C = \sqrt{\frac{1}{K_{c_1}}} \times K_{c_2} $$

4. Calculation:
Substituting the given values:
$$ K_C = \sqrt{\frac{1}{2.4 \times 10^{30}}} \times 1.4 $$
$$ K_C = \frac{1.4}{\sqrt{2.4 \times 10^{30}}} $$
$$ K_C = \frac{1.4}{1.55 \times 10^{15}} $$
$$ K_C = 9.03 \times 10^{-16} $$

Final Answer:
The equilibrium constant for the reaction is $9.0 \times 10^{-16}$.

Answer 2

The given reactions and their equilibrium constants are:

1. \( 2\text{NO}(g) \rightleftharpoons \text{N}_2(g) + \text{O}_2(g); K_1 = 2.4 \times 10^{20} \)

2. \( \text{NO}(g) + \frac{1}{2}\text{Br}_2(g) \rightleftharpoons \text{NOBr}(g); K_2 = 1.4 \)

We need to calculate the equilibrium constant \( K_C \) for the target reaction:

\( \frac{1}{2}\text{N}_2(g) + \frac{1}{2}\text{O}_2(g) + \frac{1}{2}\text{Br}_2(g) \rightleftharpoons \text{NOBr}(g) \)

To find \( K_C \), we'll combine the given reactions:

1. Reverse reaction 1 and divide it by 2, since we need \( \frac{1}{2}\text{N}_2 \) and \( \frac{1}{2}\text{O}_2 \):

\( \text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{NO}(g); K_1' = \frac{1}{K_1} = \frac{1}{2.4 \times 10^{20}} \)

Divide by 2:

\( \frac{1}{2}\text{N}_2(g) + \frac{1}{2}\text{O}_2(g) \rightleftharpoons \text{NO}(g); K_3 = \sqrt{K_1'} \)

Thus,

\( K_3 = \sqrt{\frac{1}{2.4 \times 10^{20}}} = \frac{1}{\sqrt{2.4 \times 10^{20}}} \)

2. Combine the new equation with reaction 2:

The equilibrium constant for the target reaction:

\( K_C = K_3 \times K_2 \)

Since: \( K_3 = \frac{1}{\sqrt{2.4 \times 10^{20}}} \)

Then,

\( K_C = \frac{1}{\sqrt{2.4 \times 10^{20}}} \times 1.4 \)

Calculating \( \sqrt{2.4 \times 10^{20}} \):

\( \sqrt{2.4 \times 10^{20}} = 1.55 \times 10^{10} \)

Thus,

\( K_C = \frac{1.4}{1.55 \times 10^{10}} = 9.0 \times 10^{-16} \)

Therefore, the equilibrium constant \( K_C \) for the given reaction is \( 9.0 \times 10^{-16} \).