Question

At a given temperature K$_1$, K$_2$ and K$_3$ are the equilibrium constants for the following reactions 1, 2 and 3 respectively:
CH$_4$(g) + H$_2$O(g) = CO(g) + 3H$_2$(g) ___(1)
CO(g) + H$_2$O(g) = CO$_2$(g) + H$_2$(g) ___(2)
CH$_4$(g) + 2H$_2$O(g) = CO$_2$(g) + 4H$_2$(g) ___(3)
Then the K$_1$, K$_2$ and K$_3$ are related as

• A. K$_3$ = (K$_1$ K$_2$)$^{0.5}$
• B. K$_3$ = K$_1$ + K$_2$
• C. K$_3$ = K$_1$ K$_2$
• D. K$_3$ = (K$_1$ K$_2$)$^2$

Hint:

Remember the rules for manipulating equilibrium constants:
1. If reactions are added, multiply their equilibrium constants (\( K_{\text{new}} = K_1 \cdot K_2 \)).
2. If a reaction is reversed, take the reciprocal (\( K_{\text{new}} = 1/K \)).
3. If a reaction is multiplied by a factor \( n \), raise the constant to that power (\( K_{\text{new}} = K^n \)).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks for the mathematical relationship between the equilibrium constants (\( K_1, K_2, K_3 \)) of three related chemical reactions.
This is a standard problem in chemical reaction equilibrium.

Step 2: Key Formula or Approach:
For any chemical reaction, the equilibrium constant is defined as the product of the activities (or partial pressures for ideal gases) of the products divided by those of the reactants, raised to their stoichiometric coefficients.
If a reaction (3) is the algebraic sum of reaction (1) and reaction (2):
\[ \text{Reaction (3)} = \text{Reaction (1)} + \text{Reaction (2)} \]
Then its equilibrium constant is the product of the individual equilibrium constants:
\[ K_3 = K_1 \cdot K_2 \]

Step 3: Detailed Explanation:

• Write the equilibrium constant expressions for each reaction:
\[ K_1 = \frac{P_{\text{CO}} \cdot P_{\text{H}_2}^3}{P_{\text{CH}_4} \cdot P_{\text{H}_2\text{O}}} \]
\[ K_2 = \frac{P_{\text{CO}_2} \cdot P_{\text{H}_2}}{P_{\text{CO}} \cdot P_{\text{H}_2\text{O}}} \]
\[ K_3 = \frac{P_{\text{CO}_2} \cdot P_{\text{H}_2}^4}{P_{\text{CH}_4} \cdot P_{\text{H}_2\text{O}}^2} \]

• Multiply \( K_1 \) and \( K_2 \) together:
\[ K_1 \cdot K_2 = \left( \frac{P_{\text{CO}} \cdot P_{\text{H}_2}^3}{P_{\text{CH}_4} \cdot P_{\text{H}_2\text{O}}} \right) \cdot \left( \frac{P_{\text{CO}_2} \cdot P_{\text{H}_2}}{P_{\text{CO}} \cdot P_{\text{H}_2\text{O}}} \right) \]

• Cancel the common term \( P_{\text{CO}} \) from the numerator and denominator:
\[ K_1 \cdot K_2 = \frac{P_{\text{CO}_2} \cdot P_{\text{H}_2}^4}{P_{\text{CH}_4} \cdot P_{\text{H}_2\text{O}}^2} \]

• Notice that this expression is identical to the equilibrium constant for the third reaction:
\[ K_1 \cdot K_2 = K_3 \]


Step 4: Final Answer:
The equilibrium constants are related by \( K_3 = K_1 K_2 \).