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

Consider the following reaction: CO ( g ) + 2 H 2 ( g ) ⇌ CH 3 OH ( g ) Initial moles: CO: 0.6 H₂: a CH₃OH: 0 At equilibrium (Eqm): CO: 0.4 H₂: a − 0.4 CH₃OH: 0.2 Using the ideal gas law, P V = n RT : 24 × 2.463 = ( a + 0.2 ) × 0.0821 × 900 Solving for a , we find a = 0.6 . The equilibrium constant K P is calculated as: K P = 0.4 × ( 0.2 ) 2 0.2 × ( 0.8 24 ) − 2 = 1.389 × 1 0 − 2 atm − 2

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Passage / Comprehension

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:

1.39

Solution:

Consider the following reaction:

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

CO: $0.6$

H₂: $a$

CH₃OH: $0$

At equilibrium (Eqm):

CO: $0.4$

H₂: $a - 0.4$

CH₃OH: $0.2$

Using the ideal gas law, $P V = n RT$ :

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

Solving for $a$ , we find $a = 0.6$ .

The equilibrium constant $K_{P}$ is calculated as:

$K_{P} = \frac{0.2}{0.4 \times ( 0.2 ) ^{2}} \times (\frac{24}{0.8})^{- 2} = 1.389 \times 1 0^{- 2} atm^{- 2}$

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

⏱ 2mℹ️ Source: QB

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