Hydrogen (a moles ) and iodine (b moles ) react to give 2x moles of the HI at equilibrium . The total number of moles at equilibrium is

Options

A. a+b+2x
B. (a-b)+(b-2x)
C. (a+b)
D. a+b-x

Answer

To solve the problem, we need to analyze the reaction between hydrogen (H2) and iodine (I2) to form hydrogen iodide (HI). The reaction can be represented as follows:

\[ \text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI} \]

### Step-by-step Solution:

1. **Identify Initial Moles:**
- Initially, we have:
- Moles of H2 = \( a \)
- Moles of I2 = \( b \)
- Moles of HI = \( 0 \)

2. **Define Change in Moles:**
- Let \( x \) be the amount of H2 and I2 that reacts. According to the stoichiometry of the reaction:
- Moles of H2 at equilibrium = \( a - x \)
- Moles of I2 at equilibrium = \( b - x \)
- Moles of HI at equilibrium = \( 2x \)

3. **Calculate Total Moles at Equilibrium:**
- The total number of moles at equilibrium can be calculated by adding the moles of all species present at equilibrium:
\[ \text{Total moles at equilibrium} = \text{Moles of H2} + \text{Moles of I2} + \text{Moles of HI} \]
- Substituting the values we found:
\[ \text{Total moles at equilibrium} = (a - x) + (b - x) + 2x \]

4. **Simplify the Expression:**
- Now, simplify the expression:
\[ \text{Total moles at equilibrium} = a - x + b - x + 2x = a + b - 2x + 2x \]
- This simplifies to:
\[ \text{Total moles at equilibrium} = a + b \]

### Final Answer:
The total number of moles at equilibrium is \( a + b \).

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