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
    Commented May 27, 2020 at 11:32

1 Answer 1


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:

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


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