# Finding the error in attempting the equilibrium question through isotherm equations

Consider the partial decomposition of A as:

$$\ce{2A(g) <=> 2B(g) + C(g)}$$

At equilibrium $$\pu{700 ml}$$ of gaseous mixture contains $$\pu{100 ml}$$ of gas C at $$\pu{10 atm}$$ and $$\pu{300 K}$$. What is the value of $$\mathrm{K_p}$$ for this reaction?

Attempt:

$$\ce{2A<=> 2B + C}$$

Let $$\ce{n_A}= 2x - 2y$$

$$\implies \ce{n_B = 2y}, \ce{n_C = y}$$

$$\dfrac{n_x}{n_{total}}= \dfrac{P_x}{P_{total}}= \dfrac{V_x}{V_{total}}$$

For C,

$$\implies \dfrac{y}{2x+y}= 100/700 = 1/7$$

$$\implies x = 3y$$

Again for see, from pressure relation

$$y/ (7y) = \dfrac {10}{P_t}$$

$$\implies P_t = \pu{ 70 atm}$$

Now, $$\mathrm{K_p} = \dfrac{(p_C)(p_B)^2}{p_A^2}$$

$$\implies K_p = \dfrac{10 \times (2y/7y \times 70)^2}{(4y/7y \times 70)^2} = 10/ 4 = 5/2$$

But answer given is $$10/ 28$$.

Please let me know my error.

While Solving this, you have assumed that the partial pressure is $$10$$ atm., which is not the case here. The question actually says that the total pressure ($$P_t$$) is $$10$$ atm.
It will be very much easier if you do the whole calculation by considering partial pressures instead of no. of moles. Suppose the initial pressure was $$P^0$$, which is nothing but the initial pressure of $$A$$. Now, after attaining equilibrium let the new partial pressure of $$A$$ becomes $$P^0 -P$$. So, partial pressure of $$B$$ will be $$P$$, and that of $$C$$ will be $$\frac{P}{2}$$.
Thus ,total pressure ($$P_t$$)= $$P^0 + \frac{P}{2} = 10$$ and by the given volume condition, $$\frac{\frac{P}{2}}{P^0 + \frac{P}{2}} = \frac{1}{7}$$.
Solving these two equations will give, $$P = \frac{20}{7}$$atm. and $$P^0 = \frac{60}{7}$$ atm. Thus, new partial pressure of $$A$$ will be = $$P^0 -P = \frac{40}{7}$$ atm. and partial pressure of $$B$$= $$P = \frac{20}{7}$$atm. and that of $$C$$ ($$\frac{P}{2}$$) = $$\frac{10}{7}$$atm.
So, $$K_p = \frac{p_C \times p_B^2}{p_A^2} = \frac{\frac{P}{2} \times P^2}{(P^0-P)^2} = \frac{10}{28}$$ Thus, the assumption of total pressure as the partial pressure created the error.