Chemical Equilibrium: Conceptual Doubts Cleared (4)

Let $x$ moles of $N_2$ react. Then $3x$ moles of $H_2$ react and $2x$ moles of $NH_3$ form.

In molarity (volume = 2 L):

Wait — let me redo with cleaner variables. Let $a$ = molarity of $N_2$ that reacts. Then $H_2$ reacted = $3a$, $NH_3$ formed = $2a$.

$K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(2a)^2}{(0.5 - a)(1.5 - 3a)^3} = 0.5$

Note that $1.5 - 3a = 3(0.5 - a)$. Let $y = 0.5 - a$:

$0.5 = \frac{(2a)^2}{y \cdot 27 y^3} = \frac{4a^2}{27 y^4}$

$13.5 y^4 = 4a^2$

Try $a = 0.1$: $y = 0.4$, $y^4 = 0.0256$, $13.5 \times 0.0256 = 0.3456$. RHS: $4 \times 0.01 = 0.04$. LHS too big.

Try $a = 0.3$: $y = 0.2$, $y^4 = 0.0016$, $13.5 \times 0.0016 = 0.0216$. RHS: $4 \times 0.09 = 0.36$. RHS now bigger.

Try $a = 0.2$: $y = 0.3$, $y^4 = 0.0081$, $13.5 \times 0.0081 = 0.10935$. RHS: $4 \times 0.04 = 0.16$. RHS bigger.

Try $a = 0.18$: $y = 0.32$, $y^4 = 0.01049$, $13.5 \times 0.01049 = 0.1416$. RHS: $4 \times 0.0324 = 0.1296$. Very close.

Try $a = 0.185$: $y = 0.315$, $y^4 \approx 0.00984$, LHS $\approx 0.1329$. RHS $\approx 0.1369$. Even closer.

Take $a \approx 0.184$, so $[NH_3] = 2a \approx 0.368\,\text{M}$.

Final: $[NH_3] \approx 0.37$ M.