At equilibrium, 3.00 mol of HI gas are found. How many moles of H2 and I2 are present at equilibrium? (Give your answer to two decimal places.)
For the reaction, H2 + I2 <—-> 2HI the initial mixture contains 6.50 mol H2, 8.30 mol I2, and no moles of HI. At equilibrium, 3.00 mol of HI gas are found. How many moles of H2 and I2 are present at equilibrium? (Give your answer to two decimal places.)
The Correct Answer and Explanation is:
To solve this equilibrium problem, we’ll use an ICE table (Initial, Change, Equilibrium) to track the changes in the amounts of reactants and products as the reaction proceeds.
Step 1: Write the balanced chemical equation
H2(g)+I2(g)⇌2HI(g)H_2(g) + I_2(g) \rightleftharpoons 2HI(g)H2(g)+I2(g)⇌2HI(g)
Step 2: Define initial moles and changes
From the problem, we know the initial moles and the amount of HI at equilibrium.
- Initial moles:
- H2 = 6.50 mol
- I2 = 8.30 mol
- HI = 0 mol
- At equilibrium:
- HI = 3.00 mol
Step 3: Set up the ICE table
| H2 (mol) | I2 (mol) | HI (mol) | |
|---|---|---|---|
| Initial | 6.50 | 8.30 | 0 |
| Change | -x | -x | +2x |
| Equilibrium | 6.50 – x | 8.30 – x | 3.00 |
Step 4: Use the equilibrium concentration of HI to find x
Since 2 moles of HI are produced for every 1 mole of H2 and I2 that react, the change in moles of HI will be 2x. At equilibrium, 3.00 mol of HI is present, so we can set up the equation:2x=3.00 mol2x = 3.00 \, \text{mol}2x=3.00mol
Solving for xxx:x=3.002=1.50 molx = \frac{3.00}{2} = 1.50 \, \text{mol}x=23.00=1.50mol
Step 5: Calculate the equilibrium amounts of H2 and I2
Now that we know x=1.50x = 1.50x=1.50 mol, we can substitute this into the expressions for the equilibrium amounts of H2 and I2.
- H2 at equilibrium: 6.50−1.50=5.00 mol6.50 – 1.50 = 5.00 \, \text{mol}6.50−1.50=5.00mol
- I2 at equilibrium: 8.30−1.50=6.80 mol8.30 – 1.50 = 6.80 \, \text{mol}8.30−1.50=6.80mol
Final Answer:
At equilibrium, the moles of each substance are:
- H2: 5.00 mol
- I2: 6.80 mol
- HI: 3.00 mol (given in the problem)
Thus, the number of moles of H2 and I2 at equilibrium are 5.00 mol and 6.80 mol, respectively.
