Question

At 450⁰C, Kp =6.5 x 10^-3 for the ammonia synthesis reaction. Assume that a reaction vessel with a movable piston initially contains 3.0 moles of H₂(g) and 1.0 moles of N₂(g). Make a plot to show how the partial pressure of NH₃(g) present at equilibrium varies for the total pressures of 1.0 atm, 10.0 atm, and 100. atm, and 1000. atm (assuming that Kp remains constant). [Note: Assume these total pressures represent the initial total pressure of H₂(g) plus N₂(g), where P_NH3 = 0.]

GIVEN: Kp = 6.5 X 10^-3

Short answer

increases, and a larger fraction of N₂ and H₂ is converted to NH₃.

That is, as ${\mathrm{P}}_{\mathrm{total}}$increases (V decreases), the reaction shifts further to the right, as predicted by Le Chatelier’s principle.

Step-by-step solution

First we have to calculate partial pressure of NH3 at different pressures:

At 1.0 atm: ${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{NH}}_3\left(\mathrm{g}\right)$

Initial pressure: 0.25 atm + 0.75 atm →0

At equilibrium:(0.25 –x)+ (0.75 – 3x)→2x

$\frac{{\left(2\mathrm{x}\right)}^2}{{(0.75-3\mathrm{x})}^2(0.25-\mathrm{x})}=6.5\times {10}^{-3}$

By using successive approximation:

$\mathbf{x}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{atm}\mathbf{;}{\mathbf{P}}_{{\mathbf{NH}}_{\mathbf{3}}}\mathbf{=}\mathbf{2}\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{024}\mathbf{atm}$

At 10atm ${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{NH}}_3\left(\mathrm{g}\right)$

Initial pressure: 2.5 atm 7.5 atm 0

Equilibrium: 2.5 –$x$ 7.5 − 3$x$2 $x$

$\frac{{\left(2\mathrm{x}\right)}^2}{{(7.5-3\mathrm{x})}^2(2.5-\mathrm{x})}=6.5\times {10}^{-3}$

using successive approximations:

$\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{69}\mathbf{atm}\mathbf{;}{\mathbf{P}}_{{\mathbf{NH}}_{\mathbf{3}}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{4}\mathbf{atm}$

At 100 atm${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{NH}}_3\left(\mathrm{g}\right)$

Initial pressure: 25 atm + 75 atm → 0

Equilibrium: (25 – x)+(75 – 3x)→2x

$\frac{{\left(2\mathrm{x}\right)}^2}{{(75-3\mathrm{x})}^3(2.5-\mathrm{x})}=6.5\times {10}^{-3}$

Solving by successive approximations:

$\mathbf{x}\mathbf{=}\mathbf{16}\mathbf{atm}\mathbf{;}{\mathbf{P}}_{{\mathbf{NH}}_{\mathbf{3}}}\mathbf{=}\mathbf{32}\mathbf{atm}$

At 1000 atm : ${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{NH}}_3\left(\mathrm{g}\right)$

Initial pressure 250atm + 750 atm → 0

Suppose 250atm N₂ react completely, then

New initial pressure 0 + 0 5.0 x 10²

At equilibrium:x+x 5.0 x 10²–2x

$\frac{{(5.0\times {10}^2-2\mathrm{x})}^2}{{\left(3\mathrm{x}\right)}^3\left(\mathrm{x}\right)}=6.5\times {10}^{-3}$

Using successive approximations:

$\mathrm{x}=32\mathrm{atm};{\mathrm{P}}_{{\mathrm{NH}}_3}=5.0\times {10}^2\mathrm{atm}-2\mathrm{x}=440\mathrm{atm}$

Now plot the graph between: logPNH3Vs logPtotal

The results are plotted as log $P_{N{H}_3}$ versus log $P_{total}$. Notice that as $p_{total}$ increases, a larger fraction of N₂ and H₂ is converted to NH₃, that is, as $p_{total}$ increases (V decreases), the reaction shifts further to the right, as predicted by Le Chatelier’s principle.