Question

Liquid nitrogen trichloride is heated in a 2.50−L closed reaction vessel until it decomposes completely to gaseous elements. The resulting mixture exerts a pressure of 818 mmHg at 95°C. What is the partial pressure of each gas in the container?

Answer 1

`p_(N_2)=204.5mmHg\\p_(Cl_2)=613.5mmHg`

Step-by-step explanation:

Hello,

In this case, the undergoing chemical reaction is:

`2NCl_3(g)\rightarrow N_2(g)+3Cl_2(g)`

Thus, by knowing that the nitrogen trichloride is completely decomposed, one assumes there is one mole of nitrogen and three moles of chlorine (stoichiometric coefficients) as a basis to compute the partial pressures since they have the mole ratio from the nitrogen trichloride. Hence, the mole fractions result:

`x_(N_2)=(1)/(1+3)=0.25\\ x_(Cl_2)=1-0.25=0.75`

In such a way, for the final pressure 818 mmHg, the partial pressures become:

`p_(N_2)=x_(N_2)p_T=0.25*818mmHg=204.5mmHg\\p_(Cl_2)=x_(Cl_2)p_T=0.75*818mmHg=613.5mmHg`

Best regards.

Answer 2

1. Partial pressure of N2 is 204.5 mmHg

2. Partial pressure of Cl2 is 613.5 mmHg

Step-by-step 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