Question

3. Consider the ammonia production reaction, N2(g)+3H2(g) = 2NH3 (g) The equi librium constant for this reaction at 298K is 6.10x 105. When the system starts with 2 mol of NH3 (no H2 or N2), it reaches the equilibrium at the total pressure of 2.0 bar (a) What is the mole fraction of each species at equilibrium? (b) If we increase the total pressure to 3.0 bar, will it be stimulating the decompo- sition of ammonia? Explain

Answer 1

At equilibrium, we know that the sum of all partial pressures is 2 bar:

$p=p_{NH_{3}}+p_{H_{2}}+p_{N_{2}}=2.0bar$

We also know that the pressure if hydrogen will be three times the pressure of nitrogen, since they are not initially present and they are generated in a 1:3 ratio:

$p_{NH_{3}}+4p_{N_{2}}=2.0bar$ (equation 1)

We can use these unknown values to write the constant of equilibrium:

$6.10x10^{5}=\frac{(p_{NH_{3}})^{2}}{p_{N_{2}}\cdot (3p_{N_{2}})^{3}}=\frac{(p_{NH_{3}})^{2}}{27(p_{N_{2}})^{4}}$

If we apply the square root to both sides of the equation, we get:

$781=\frac{p_{NH_{3}}}{\sqrt{27}(p_{N_{2}})^{2}}$

We rearrange this:

$p_{NH_{3}}=781\cdot \sqrt{27}(p_{N_{2}})^{2}=4058(p_{N_{2}})^{2}$

We can replace this into equation 1:

$4058(p_{N_{2}})^{2}+4p_{N_{2}}=2.0$

Which is a quadratic equation which can be solved to yield:

$x_{1}=0.022;x_{2}=-0.022$

Since pressures can only have positive values, x₁ is the answer.

This means that the pressures in equilibrium are:

N₂ = 0.022 bar; H₂ = 0.066 bar; NH₃ = 1.912 bar

Since the partial pressures are equal to the product of the molar fraction by the total pressure, the molar fractions are:

$N_{2}:x=\frac{p_{N_{2}}}{p_{Total}}=\frac{0.022bar}{2.0bar}=0.011$

$H_{2}:x=\frac{p_{H_{2}}}{p_{Total}}=\frac{0.066bar}{2.0bar}=0.033$

$NH_{3}:x=\frac{p_{NH_{3}}}{p_{Total}}=\frac{1.912bar}{2.0bar}=0.956$

b) If the pressure is increased, the system will shift towards the side that counteracts the pressure increase (as stated by Le Chatelier's principle). In this case, this means a shift towards the production of more ammonia, since in this direction the total number of gaseous moles is decreased (4 moles of gas react to obtain only 2), which generates a decrease in the pressure.