When an amount ammonia is added at $\pu{600 K}$ in a $\pu{1 L}$ container the following reaction takes place:

$$\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)}$$

The equilibrium constant $K_c = 4.20$ at $\pu{600 K}.$

At equilibrium, it is known that $\pu{0.200 mol}$ of $\ce{N2}$ gas exist in the container. What amount of ammonia was added at the start of the reaction? Choose from the answers below:

$\pu{0.826 mol};$ $\pu{0.482 mol};$ $\pu{1.226 mol};$ $\pu{0.400 mol};$ $\pu{0.800 mol}.$

My attempt

I created an ICE table, but I think it's wrong:

$$ \begin{array}{lccc} \ce{&N2(g) &+ &3 H2(g) &<=> &2 NH3(g)} \\ \text{I} & 0 && 0 && y \\ \text{C} & +x && +3x && -4x \\ \text{E} & 0.2 && 0.6 && y-0.8 \\ \end{array} $$

$$K_c = \frac{(y - 0.8)^2}{0.200\times 0.6^3}$$

Solving the equation with Maple gives me none of the answers above. I must have messed up on the table. Can anyone tell me where I messed up?

  • 2
    $\begingroup$ The method is ok, but you made a slip, the term for ammonia is not -4x, correct this and you will get one of the answers. In an exam you can get the answer easily with a hand calculator. $\endgroup$ – porphyrin May 27 at 11:32
  • $\begingroup$ Thank you very much. $\endgroup$ – Carl May 27 at 11:45

Since the volume is given and it's a constant, the initial amount of ammonia $n_0(\ce{NH3})$ can be found from its initial concentration $c_0(\ce{NH3}):$

$$n_0(\ce{NH3}) = c_0(\ce{NH3})\times V\tag{1}$$

To find $c_0(\ce{NH3}),$ an ICE table might indeed come in handy; however, yours needs corrections. First, I suggest to rewrite it according to the process occurring in the system, namely dissociation of ammonia:

$$ \begin{array}{lccc} \ce{&2 NH3(g) &<=> &N2(g) &+ &3 H2(g)} \\ \text{I} & c_0 && 0 && 0 \\ \text{C} & -2x && +x && +3x \\ \text{E} & c_0 - 2x && x && 3x \\ \end{array} $$

Second, note that the provided equilibrium constant $K_c$ is given for the synthesis of ammonia, hence in our case a reciprocal value is relevant:

$$ \begin{align} \frac{1}{K_c} &= \frac{[\ce{N2}][\ce{H2}]^3}{[\ce{NH3}]^2}\tag{2.1}\\ \frac{1}{K_c} &= \frac{x (3x)^3}{(c_o - 2x)^2}\tag{2.2}\\ \frac{1}{K_c} &= \frac{27x^4}{(c_o - 2x)^2}\tag{2.3}\\ \frac{1}{\sqrt{K_c}} &= \frac{3\sqrt{3}x^2}{c_o - 2x}\tag{2.4} \end{align} $$

$$c_0 = 3\sqrt{3K_c}x^2 + 2x\tag{3}$$

At the equilibrium there is $\pu{0.200 mol}$ of nitrogen in the $\pu{1 L}$ vessel, so $x = \pu{0.200 mol L-1},$ and the initial amount of ammonia can be found as follows:

$$ \begin{align} n_0(\ce{NH3}) &= (3\sqrt{3K_c}x^2 + 2x) × V\\ &= (3\sqrt{3\times 4.20}\times (\pu{0.200 mol L-1})^2 + 2\times \pu{0.200 mol L-1})\times \pu{1 L} \\ &= \pu{0.826 mol}\tag{4} \end{align} $$

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.