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

C(s) + CO2(g) ⇌ 2CO (g)

Calculate Kc for this reaction at the above temperature

Asked by Abhisek | 1 year ago |  77

##### Solution :-

Let us assume that the solution is 100g in total.

Given, mass of CO = 90.55 g

Now, the mass of CO2 = (100 – 90.55) = 9.45 g

Now, number of moles of CO,

$$n_{CO}=\dfrac{90.5}{28}=3.234\: mol$$

Number of moles of CO2

$$n_{CO_{2}}=\dfrac{9.45}{44}=0.215\: mol$$

Partial pressure of CO,

$$PCO= \dfrac{n_{CO}}{n_{CO}+n_{CO_{2}}}\times p_{total}$$

$$=\dfrac{3.234}{3.234+0.215}\times 1=0.938$$

Partial pressure of CO2

$$P_{CO_{2}}=\dfrac{n_{CO_{2}}}{n_{CO}+n_{CO_{2}}}\times p_{total}$$

$$=\dfrac{0.215}{3.234+0.215}\times 1=0.062, atm$$

Therefore,

$$Kp= \dfrac{[CO]^{2}}{[CO_{2}]}\\ \\$$

$$=\dfrac{(0.938)^{2}}{0.062}\\ \\$$

$$=14.19$$

For the given reaction,

∆n = 2 – 1 = 1

We know that,

$$Kp = Kc(RT) ^{\Delta n}$$

$$\Rightarrow 14.19 = K_{c}(0.082\times 1127)^{1}\\ \\ \Rightarrow K_{c}$$

$$=\dfrac{14.19}{0.082\times 1127}\\ \\$$

$$=0.154(approximately)​$$

Answered by Pragya Singh | 1 year ago

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