Question

Consider the following reactions in which all the reactants and products are present in gaseous state:
\(2xy \rightleftharpoons x_2 + y_2\) \(K_1 = 2.5 \times 10^5\)
\(xy + \frac{1}{2}z_2 \rightleftharpoons xyz\) \(K_2 = 5 \times 10^{-3}\)
The value of \(K_3\) for the equilibrium \(\frac{1}{2}x_2 + \frac{1}{2}y_2 + \frac{1}{2}z_2 \rightleftharpoons xyz\) is:

• A. \(2.5 \times 10^{-3}\)
• B. \(2.5 \times 10^3\)
• C. \(1.0 \times 10^{-5}\)
• D. \(5 \times 10^{-3}\)

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We must algebraically manipulate the given equilibrium equations to yield the target equation. When adding/subtracting equations or multiplying by constants, the equilibrium constants undergo multiplication/division or exponentiation, respectively.
Step 2: Key Formula or Approach:
1. Reversing an equation: $K_{new} = \frac{1}{K_{old}}$.
2. Multiplying an equation by a factor $n$: $K_{new} = (K_{old})^n$.
3. Adding two equations: $K_{new} = K_{eq1} \times K_{eq2}$.
Step 3: Detailed Explanation:
We are given:
Eq 1: $2xy \rightleftharpoons x_2 + y_2$ with $K_1 = 2.5 \times 10^5$.
Eq 2: $xy + \frac{1}{2}z_2 \rightleftharpoons xyz$ with $K_2 = 5 \times 10^{-3}$.
Target Eq 3: $\frac{1}{2}x_2 + \frac{1}{2}y_2 + \frac{1}{2}z_2 \rightleftharpoons xyz$ with $K_3$.
Notice that the target equation has $x_2$ and $y_2$ on the reactant side, whereas Eq 1 has them on the product side. Also, the coefficients are $1/2$ instead of $1$.
First, reverse Eq 1 to get $x_2$ and $y_2$ as reactants:
$x_2 + y_2 \rightleftharpoons 2xy$
The new equilibrium constant is $K_1' = \frac{1}{K_1}$.
Next, divide this reversed equation by 2 (multiply by 1/2) to match the coefficients in the target equation:
$\frac{1}{2}x_2 + \frac{1}{2}y_2 \rightleftharpoons xy$
The new equilibrium constant is $K_1'' = (K_1')^{1/2} = \sqrt{\frac{1}{K_1}}$.
Calculate $K_1''$:
$K_1'' = \frac{1}{\sqrt{2.5 \times 10^5}} = \frac{1}{\sqrt{25 \times 10^4}} = \frac{1}{5 \times 10^2} = \frac{1}{500} = 0.002 = 2 \times 10^{-3}$.
Now, add this manipulated Eq 1 to Eq 2:
(Manipulated Eq 1): $\frac{1}{2}x_2 + \frac{1}{2}y_2 \rightleftharpoons xy$
(Eq 2): $xy + \frac{1}{2}z_2 \rightleftharpoons xyz$
When we add them, the intermediate $xy$ cancels out:
$\frac{1}{2}x_2 + \frac{1}{2}y_2 + \frac{1}{2}z_2 \rightleftharpoons xyz$. This is exactly the target equation!
When adding equations, multiply their equilibrium constants:
$K_3 = K_1'' \times K_2 = (2 \times 10^{-3}) \times (5 \times 10^{-3})$.
$K_3 = 10 \times 10^{-6} = 1.0 \times 10^{-5}$.
Step 4: Final Answer:
The value of $K_3$ is $1.0 \times 10^{-5}$.