<u>Answer:</u> The moles of bromine gas at equilibrium is 0.324 moles.
<u>Explanation:</u>
To calculate the molarity of solution, we use the equation:
.......(1)
Calculating the initial moles of hydrogen and bromine gas:
Moles of hydrogen gas = 0.682 mol
Volume of solution = 2.00 L
Putting values in equation 1, we get:
![\text{Molarity of solution}=\frac{0.682mol}{2.00L}=0.341M](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%20of%20solution%7D%3D%5Cfrac%7B0.682mol%7D%7B2.00L%7D%3D0.341M)
Moles of bromine gas = 0.440 mol
Volume of solution = 2.00 L
Putting values in equation 1, we get:
![\text{Molarity of solution}=\frac{0.440mol}{2.00L}=0.220M](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%20of%20solution%7D%3D%5Cfrac%7B0.440mol%7D%7B2.00L%7D%3D0.220M)
Now, calculating the molarity of hydrogen gas at equilibrium by using equation 1:
Equilibrium moles of hydrogen gas = 0.566 mol
Volume of solution = 2.00 L
Putting values in equation 1, we get:
![\text{Molarity of solution}=\frac{0.566mol}{2.00L}=0.283M](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%20of%20solution%7D%3D%5Cfrac%7B0.566mol%7D%7B2.00L%7D%3D0.283M)
Change in concentration of hydrogen gas = 0.341 - 0.283 = 0.058 M
This change will be same for bromine gas.
Equilibrium concentration of bromine gas = ![(\text{Initial concentration}-\text{Change in concentration})=0.220-0.058=0.162M](https://tex.z-dn.net/?f=%28%5Ctext%7BInitial%20concentration%7D-%5Ctext%7BChange%20in%20concentration%7D%29%3D0.220-0.058%3D0.162M)
Now, calculating the moles of bromine gas at equilibrium by using equation 1:
Molarity of bromine gas = 0.162 M
Volume of solution = 2.00 L
Putting values in equation 1, we get:
![0.162M=\frac{\text{Moles of bromine gas}}{2.00L}\\\\\text{Moles of bromine gas}=(0.162mol/L\times 2.00L)=0.324mol](https://tex.z-dn.net/?f=0.162M%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20bromine%20gas%7D%7D%7B2.00L%7D%5C%5C%5C%5C%5Ctext%7BMoles%20of%20bromine%20gas%7D%3D%280.162mol%2FL%5Ctimes%202.00L%29%3D0.324mol)
Hence, the moles of bromine gas at equilibrium is 0.324 moles.