Ncert Exercises & Solutions

7.23 At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2 in equilibrium with soild carbon has 90.55% CO by mass

C (s) + CO2 (g) 2CO (g)

Calculate Kc for this reaction at the above temperature.

Let the total mass of the gaseous mixture be 100 g.
Mass of CO = 90.55 g
And, mass of ${CO}_{2}$ = (100 – 90.55) = 9.45 g

Now, the number of moles of CO, $n_{co} = \frac{90.55}{28} = 3.234 mol$

Number of moles of ${CO}_{2}, n_{{co}_{2}} = \frac{9.45}{44} = 0.215 mol$

The partial pressure of CO,

$p_{co} = \frac{n_{co}}{n_{co} + n_{{co}_{2}}} \times p_{total}$
$= \frac{3.234}{3.234 + 0.215} \times 1$
$= 0.938 atm$

The partial pressure of ${CO}_{2}$ ,

$p_{{co}_{2}} = \frac{n_{{co}_{2}}}{n_{co} + n_{{co}_{2}}} \times p_{total}$
$= \frac{0.215}{3.234 + 0.215} \times 1$
$= 0.062 atm$

Therefore $K_{p}$ =

$= \frac{{[CO]}^{2}}{[{CO}_{2}]}$
$= \frac{{(0.938)}^{2}}{0.062}$
$= 14.19$

For the given reaction,

$∆ n = 2 - 1 = 1$
$We know that,$
$K_{p} = K_{c} {(RT)}^{∆ n}$
$\Rightarrow 14.19 = K_{c} {(0.082 \times 1127)}^{1}$
$\Rightarrow K_{c} = \frac{14.19}{0.082 \times 1127}$
$= 0.154 (approx .)$

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