Determine the value of K for the reaction,
2NOCl(g) equilibrium reaction arrow O2(g) + N2(g) + Cl2(g)
by combining the equations below to give the needed equation and using their K values to calculate the K value of that equation.
1/2 N2(g) + 1/2 O2(g) equilibrium reaction arrow NO(g) with K = 0.0931
NO(g) + 1/2Cl2(g) equilibrium reaction arrow NOCl(g) with K = 0.0785

To determine the equilibrium constant (\( K \)) for the overall reaction:

\[
2NOCl(g) \rightleftharpoons O_2(g) + N_2(g) + Cl_2(g)
\]

we need to combine the two given reactions to match the desired reaction. The provided reactions and their equilibrium constants are:

1. \( \frac{1}{2} N_2(g) + \frac{1}{2} O_2(g) \rightleftharpoons NO(g) \) with \( K_1 = 0.0931 \)
2. \( NO(g) + \frac{1}{2} Cl_2(g) \rightleftharpoons NOCl(g) \) with \( K_2 = 0.0785 \)

### Step 1: Reverse the second reaction
The desired overall reaction involves the formation of \( NOCl(g) \) from \( NO(g) \), so we reverse reaction 2:

\[
NOCl(g) \rightleftharpoons NO(g) + \frac{1}{2} Cl_2(g)
\]

When a reaction is reversed, the equilibrium constant for the reverse reaction becomes the reciprocal of the original constant:

\[
K_2' = \frac{1}{K_2} = \frac{1}{0.0785} = 12.7389
\]

### Step 2: Multiply the first reaction by 2
To match the stoichiometry of the overall reaction, we multiply the first reaction by 2:

\[
N_2(g) + O_2(g) \rightleftharpoons 2NO(g)
\]

When multiplying a reaction by a factor, the equilibrium constant is raised to the power of that factor. Since we are multiplying by 2:

\[
K_1' = (K_1)^2 = (0.0931)^2 = 0.00866861
\]

### Step 3: Combine the two reactions
Now we add the two modified reactions:

1. \( N_2(g) + O_2(g) \rightleftharpoons 2NO(g) \) with \( K_1' = 0.00866861 \)
2. \( 2NOCl(g) \rightleftharpoons 2NO(g) + Cl_2(g) \) with \( K_2' = 12.7389 \)

Adding these reactions gives:

\[
2NOCl(g) \rightleftharpoons N_2(g) + O_2(g) + Cl_2(g)
\]

### Step 4: Multiply the equilibrium constants
For combined reactions, the overall equilibrium constant \( K \) is the product of the individual constants:

\[
K = K_1' \times K_2' = 0.00866861 \times 12.7389 = 0.1104
\]

### Final Answer:
The equilibrium constant \( K \) for the overall reaction is approximately \( \mathbf{0.110} \).

---

Do you want further details or explanations? Here are some related questions to deepen understanding:

1. How does reversing a reaction affect the equilibrium constant?
2. What happens to the equilibrium constant when a reaction is multiplied by a coefficient?
3. Can equilibrium constants be negative?
4. How does the stoichiometry of a reaction impact the overall equilibrium constant?
5. What role do equilibrium constants play in determining the extent of a reaction?

**Tip:** When combining equilibrium constants, remember to adjust for any changes in stoichiometry or reaction direction (reversal).

Determine the value of K for the reaction, 2NOCl(g) equilibrium reaction arrow O2(g) + N2(g) + Cl2(g) by combining the equations below to give the needed equation and using their K values to calculate the K value of that equation. 1/2 N2(g) + 1/2 O2(g) equilibrium reaction arrow NO(g) with K = 0.0931. NO(g) + 1/2Cl2(g) equilibrium reaction arrow NOCl(g) with K = 0.0785.

Learn how to calculate the equilibrium constant (K) for the reaction 2NOCl(g) → O2(g) + N2(g) + Cl2(g) by combining two given reactions and their K values. This solution involves adjusting reaction stoichiometry, reversing reactions, and multiplying equilibrium constants, using key concepts from chemical equilibrium and reaction kinetics.

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