Answer:
1. Partial pressure of N2 is 204.5 mmHg
2. Partial pressure of Cl2 is 613.5 mmHg
Explanation:
Step 1:
The equation for the reaction. This is given below:
NCl3 —> N2 (g) + Cl2 (g)
Step 2:
Balancing the equation.
NCl3 (l) —> N2 (g) + Cl2 (g)
The above equation is balanced as follow:
There are 2 atoms of N on the right side and 1 atom on the left side. It can be balance by putting 2 in front of NCl3 as shown below:
2NCl3 (l) —> N2 (g) + Cl2 (g)
There are 6 atoms of Cl on the left side and 2 atoms on the right side. It can be balance by putting 3 in front of Cl2 as shown below:
2NCl3 (l) —> N2 (g) + 3Cl2 (g)
Now the equation is balanced.
Step 2:
Determination of the mole fraction of each gas.
From the balanced equation above, the resulting mixture of the gas contains:
Mole of N2 = 1
Mole of Cl2 = 3
Total mole = 4
Therefore, the mole fraction for each gas is:
Mole fraction of N2 = mole of N2/total mole
Mole fraction of N2 = 1/4
Mole fraction of Cl2 = mole of Cl2/total mole
Mole fraction of Cl2 = 3/4
Step 3:
Determination of the partial pressure of N2.
Partial pressure = mole fraction x total pressure
Total pressure = 818 mmHg
Mole fraction of N2 = 1/4
Partial pressure of N2 = 1/4 x 818
Partial pressure of N2 = 204.5 mmHg
Step 4:
Determination of the partial pressure of Cl2.
Partial pressure = mole fraction x total pressure
Total pressure = 818 mmHg
Mole fraction of Cl2 = 3/4
Partial pressure of Cl2 = 3/4 x 818
Partial pressure of Cl2 = 613.5 mmHg
