when Br2 (5.7mol) and Cl2 (.0934mol/L)in a 61 L vessel at 604'C are allowed to come to equlibrium the mixture contains .0523 mol/L of BrCl. What concentration of Cl2 is reacted?
- i know how to find the Kc...
Br2 + Cl2 ==> 2BrCl
0.0523 mol/L BrCl at equilibrium in a 61 L vessel = 0.0523 mol/L x 61 L = mols BrCl formed at equilibrium.
1/2 x mols BrCl = mols Cl2 reacted.
Check my thinking.
Your thinking is correct so far. To find the concentration of Cl2 reacted, you need to first find the amount of BrCl formed at equilibrium.
You have already calculated the moles of BrCl formed at equilibrium using the given concentration and volume:
0.0523 mol/L x 61 L = 3.1923 mol BrCl formed at equilibrium.
Since the balanced equation shows that 1 mole of BrCl is formed for every 1/2 mole of Cl2 reacted, you need to multiply the moles of BrCl formed by 1/2 to find the moles of Cl2 reacted:
(1/2) x 3.1923 mol = 1.5961 mol Cl2 reacted.
Now, to find the concentration of Cl2 reacted, you need to divide the moles of Cl2 reacted by the volume of the vessel:
1.5961 mol / 61 L = 0.0261 mol/L.
Therefore, the concentration of Cl2 reacted in the mixture is 0.0261 mol/L.