91Ó°ÊÓ

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.

(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

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

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

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