K P of reaction, CO ( g ) + 2 H 2 ( g ) ⇌ CH 3 OH ( g ) is ____ × 1 0 − 2 atm − 2 .

CO ( g ) + 2 H 2 ( g ) ⇌ CH 3 OH ( g ) Initial 0.6 a mole Eqm 0.4 a − 0.4 0.2 PV = nRT 24 × 2.463 = ( a + 0.2 ) × 0.0821 × 900 a = 0.6 K P = 0.4 × ( 0.2 ) 2 0.2 × ( 0.8 24 ) − 2 = 1.388 × 1 0 − 2 atm − 2

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ChemistryChemical EquilibriumProblems involving KpMedium2 minQB

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An amount of 0.60 moles of CO taken in a 2.46 L vessel is maintained at 900 K along with a catalyst so that the following reaction can take place CO(g) + 2H₂ (g) ⇌ CH₃OH(g). Hydrogen is introduced until the total pressure of the system is 24 atm at equilibrium and 0.2 mole of methanol is formed.

$K_{P}$ of reaction, $CO (g) + 2 H_{2} (g) ⇌ CH_{3} OH (g)$ is $____$ $\times 1 0^{- 2} atm^{- 2}$ .

Answer:

1.38

Solution:

$CO (g) + 2 H_{2} (g) ⇌ CH_{3} OH (g)$

Initial 0.6 a mole

$Eqm 0.4 a - 0.4$ $0.2$

$PV = nRT$

$24 \times 2.463 = (a + 0.2) \times 0.0821 \times 900$

$a = 0.6$

$K_{P} = \frac{0.2}{0.4 \times ( 0.2 ) ^{2}} \times (\frac{24}{0.8})^{- 2} = 1.388 \times 1 0^{- 2} atm^{- 2}$

Stream:JEESubject:ChemistryTopic:Chemical EquilibriumSubtopic:Problems involving Kp

⏱ 2mℹ️ Source: QB

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