# Finding initial amount of ammonia using equilibrium constant and amount of nitrogen gas

### Question

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?

• 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. May 27 '20 at 11:32
• Thank you very much.
– Carl
May 27 '20 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}