Question

Be sure to answer all parts. Liquid nitrogen trichloride is heated in a 3.00−L closed reaction vessel until it decomposes completely to gaseous elements. The resulting mixture exerts a pressure of 839 mmHg at 86°C. What is the partial pressure of each gas in the container?

Answer 1

Answer:

$p_{Cl_2}=629.25mmHg\\p_{N_2}=209.75mmHg$

Explanation:

Hello,

At first, consider the chemical reaction:

$2NCl_3-->3Cl_2+N_2$

Now, considering the ideal gas equation, we compute the total moles:

$PV=nRT\\n=\frac{PV}{RT} \\n=\frac{839mmHg*\frac{1atm}{760mmHg}*3L }{0.082\frac{atm*L}{mol*K}*359K }\\ n=0.1125mol$

Then, taking into account that the total moles are stoichiometrically handed out by the half (0.05625mol), one can say that:

$n_{Cl_2}=0.05625molNCl_3*\frac{3mol Cl_2}{2molNCl_3} =0.084375molCl_2\\n_{N_2}=0.05625molNCl_3*\frac{1mol N_2}{2molNCl_3} =0.028125molN_2$

Thus, the molar fractions are:

$x_{Cl_2}=\frac{0.084375}{0.1125} =0.75\\x_{N_2}=\frac{0.028125}{0.1125} =0.25$

Finally, the partial pressures are:

$p_{Cl_2}=x_{Cl_2}P=0.75*839mmHg=629.25mmHg\\p_{N_2}=x_{N_2}P=0.25*839mmHg=209.75mmHg$

Best regards.