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 of finding the total number of moles at equilibrium when hydrogen (H2) and iodine (I2) react to form hydrogen iodide (HI), we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between hydrogen and iodine can be represented as:
\[ \text{H}_2 + \text{I}_2 \rightleftharpoons 2 \text{HI} \]

### Step 2: Define the initial moles
Let:
- Initial moles of hydrogen (H2) = \( a \) moles
- Initial moles of iodine (I2) = \( b \) moles

### Step 3: Define the change in moles
At equilibrium, we are given that there are \( 2x \) moles of HI. This means that \( x \) moles of H2 and \( x \) moles of I2 have reacted to produce \( 2x \) moles of HI.

### Step 4: Calculate the moles at equilibrium
- Moles of H2 remaining at equilibrium = \( a - x \)
- Moles of I2 remaining at equilibrium = \( b - x \)
- Moles of HI formed at equilibrium = \( 2x \)

### Step 5: Total moles at equilibrium
The total number of moles at equilibrium can be calculated by adding the remaining moles of H2, I2, and the moles of HI:
\[ \text{Total moles at equilibrium} = (a - x) + (b - x) + 2x \]

### Step 6: Simplify the equation
Now, simplify the expression:
\[ \text{Total moles at equilibrium} = a - x + b - x + 2x = a + b - 2x + 2x = a + b \]

### Conclusion
Thus, the total number of moles at equilibrium is:
\[ \text{Total moles at equilibrium} = a + b \]