Question

Question: In a 3.0-L vessel, the following equilibrium partial pressures are measured: \({{\rm{N}}_2}\),190 torr;\({{\rm{H}}_2}\), 317 torr;\({\rm{N}}{{\rm{H}}_3}\)\(1.00 \times {10^3}\)torr.

1. How will the partial pressures of\({{\rm{H}}_2},{{\rm{N}}_2}\)and \({\rm{N}}{{\rm{H}}_3}\)change if \({{\rm{H}}_2}\) is removed from the system? Will they increase, decrease, or remain the same?
2. Hydrogen is removed from the vessel until the partial pressure of nitrogen, at equilibrium, is 250 torr. Calculate the partial pressures of the other substances under the new conditions.

Short answer

a)\({{\rm{H}}_2}\)will decrease

\({\rm{N}}{{\rm{H}}_3}\)will decrease

\({{\rm{N}}_2}\)will increase

b) The partial pressures are

\({P_{{N_2}}}\)=250 torr

\({P_{{H_2}}}\)=265 torr

\({P_{{N_3}}}\)=880 torr

Step-by-step solution

Find equilibrium constant.

Given information

- The volume of a vessel is 3.0 L

-The partial pressure of \({{\rm{N}}_2}\)at equilibrium is 190 torr

\(\left( {190torr \times \frac{{1atm}}{{760torr}} = 0.250atm} \right)\)

- The partial pressure of \({\rm{N}}{{\rm{H}}_3}\)at equilibrium is 1.00 \( \times {10^3}\)torr

\(\left( {1000t{\rm{\;orr\;}} \times \frac{{1atm}}{{760torr}} = 1.316atm} \right)\)

Therefore, the equilibrium constant is

\({K_p} = \frac{{{{\left( {N{H_3}} \right)}^2}}}{{\left( {{N_2}} \right) \times {{\left( {{H_2}} \right)}^3}}}\)

\(\begin{array}{l} = \frac{{{{(1.316)}^2}}}{{0.250 \times {{(0.417)}^3}}}\\ = 95.5\end{array}\)

(a)

We have to find what would happen to the partial pressures of \({{\rm{H}}_2},{{\rm{N}}_2}\)and\({\rm{N}}{{\rm{H}}_3}\),if we remove

\({{\rm{H}}_2}\)from the system.

If we remove\({{\rm{H}}_2}\), the equilibrium will move to the left.

- The concentration of \({{\rm{H}}_2}\)will decrease.

- Since equilibrium is moved to the left, some \({\rm{N}}{{\rm{H}}_3}\)will decompose into \({{\rm{N}}_2}\) and \({{\rm{H}}_2}\),therefore, the concentration of \({\rm{N}}{{\rm{H}}_3}\) will decrease.

-And, because of decomposition of \({\rm{N}}{{\rm{H}}_3}\),the condition of \({{\rm{N}}_2}\) will decrease.

Calculate partial pressure.

(b)

After removal of hydrogen, the partial pressure of nitrogen at equilibrium is 250 torr

\(\left( {250{\rm{\;torr\;}} \times \frac{{1{\rm{ltm}}}}{{760{\rm{torr}}}} = 0.329{\rm{atm}}} \right)\)

Under new conditions:

the initial concentration of \({{\rm{H}}_2}\)is \(0.654{\rm{\;atm\;}} - x\)

the initial concentration of \({{\rm{N}}_2}\)is 0.329 atm

the initial concentration of \({\rm{N}}{{\rm{H}}_3}\) is 1.158 atm

We have to calculate the partial pressures of the other substances under the new conditions.

Now we will find the value of x,

\({K_p} = \frac{{{{\left( {{P_{N{H_3}}}} \right)}^2}}}{{{P_{{N_2}}} \times {{\left( {{P_{{H_2}}}} \right)}^3}}}\)

\(95.5 = \frac{{{{(1.158)}^2}}}{{0.329 \times {{(0.654 - x)}^3}}}\)

\(\begin{array}{l}31.42 \times {(0.654 - x)^3} = 1.341\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{(0.654 - x)^3} = 4.27 \times {10^{ - 2}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.654 - x = 0.349\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0.305{\rm{atm}}\end{array}\)

Therefore, the partial pressures of the other substances at equilibrium are

\({P_{{N_2}}} = 0.329atm \times \frac{{760{\rm{\;torr\;}}}}{{1atm}} = 250{\rm{\;torr\;}}\)

\({P_{{H_2}}} = 0.654 - x = 0.349atm \times \frac{{760{\rm{\;torr\;}}}}{{1atm}} = 265{\rm{\;torr\;}}\)

\({P_{N{H_3}}} = 1.158a{\rm{tm}} \times \frac{{760{\rm{torr}}}}{{1{\rm{atm}}}} = 880{\rm{torr}}\)