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?


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

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.

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  • $\begingroup$ Well the question doesn't clearly say that the total pressure is 10 atm, does it? $\endgroup$ – Archer Sep 29 '18 at 18:04

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