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Standard XII

Chemistry

Equilibrium Constant and Standard Free Energy Change

0.15 mol of C...

Question

0.15 mol of CO taken in a 2.5 L flask is maintained at 705 K along with a catalyst so that the following reaction takes place CO(g) + 2 $H_{2}$ (g) $⇌$ $C H_{3}$ OH(g). Hydrogen is introduced until the total pressure of the system is 8.5 atm at equilibrium and 0.08 mol of methanol is formed. $K_{p}$ and $K_{c}$ respectively are:

= 0.045 and = 12.438

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= 150.85 and = 188

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= 0.045 and = 150.85

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Solution

The correct option is C
= 0.045 and = 150.85

Initially there was 0.15 mol of CO (g) and at equilibrium 0.08 mol of CH₃OH was formed
$\begin{matrix} C O (g) & + & 2 H_{2} (g) & ⇋ & C H_{3} O H M o l e s a t s t a r & 0.15 & a & 0 M o l e s a t e q u b . & (0.15 - x) & (a - 2 x) & 0.08 o r & (0.15 - 0.08) & (a - 0.16) & 0.08 \end{matrix}$
Total moles at equilibrium $= (0.15 - 0.08) + (a - (2 \times 0.08)) + 0.08 = a - 0.01$
We can equate the above to n = number of moles at equilibrium from
n = $\frac{P V}{R T}$ P = 8.5 atm, R = 0.0821 atmL $K^{- 1}$ $m o l^{- 1}$ , V = 2.5 litre and T = 705 K
n = 0.3676
No of moles of $H_{2}$ (g) at equilibrium = n - (0.15-0.08) - 0.08 = 0.2176
Now $K_{c} = \frac{[C H_{3} O H]}{[H_{2}]^{2} [C O]}$ = 150.8585 $m o l e^{- 2}$ $l i t r e^{2}$
$K_{p}$ much more straightforward once we figure out $K_{c}$ .
$K_{p}$ = $K_{c} (R T)^{Δ n}$

= 150.85 $\times$ $(0.0821 \times 705)^{- 2}$ = 0.045 $a t m^{- 2}$

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