Can someone pls solve this question with steps?

[Kailaskumaran Anand]

 

[Sneha Duhan]

 

[deveshgarg2008]

data is missing as photo has been cut from right side

 

[Amodini]

Please find the solution attached --

*Continuation:
But this is not possible as α > 1 is unphysical.
So the maximum α = 1, implying complete dissociation.
Thus:

• Final pressure =
  
  P_COCl2=0(fully dissociated)
• P_CO=0.4, P_Cl2=0.4, P_total=0.8 atm

But the question says total pressure = 1 atm, so 0.8 atm is from the gas mixture only.

So maybe an inert gas is added to maintain total pressure at 1 atm. In that case, the extent of dissociation = 1 (i.e., α = 1), and no change in equilibrium.

BUT, the option that best fits is where α = 0.32 and final pressure of CO = 0.4 atm.

Trying:
Let α = 0.32
Then CO = 0.4 × 0.32 = 0.128
That’s too low. Try:

Let’s now solve it algebraically.

Let α = degree of dissociation

P_total=0.4(1–α)+0.4α+0.4α=0.4+0.4α=1⇒α=1.5 (invalid!)

The correct solution is when COCl₂ = 0.32, CO = 0.4, and no change in degree of dissociation, meaning equilibrium was not disturbed.

 

[sudha]

 

[oscarpiastri25]

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Let’s break down this equilibrium problem step by step.

Reaction:​

COCl2(g)⇌CO(g)+Cl2(g)\text{COCl}_2(g) \rightleftharpoons \text{CO}(g) + \text{Cl}_2(g)
We are asked to find the degree of dissociation (α) under two different conditions.

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Condition 1: Pure COCl₂ at a total equilibrium pressure of 1 atm​

1. ICE Table using partial pressures​

Let initial pressure of COCl₂ = P0P_0, and degree of dissociation = α\alpha

Species	Initial Pressure	Change	Equilibrium Pressure
COCl₂(g)	P0P_0	−P0α-P_0\alpha	P0(1−α)P_0(1 - \alpha)
CO(g)	0	+P0α+P_0\alpha	P0αP_0\alpha
Cl₂(g)	0	+P0α+P_0\alpha	P0αP_0\alpha

2. Total Equilibrium Pressure​

Ptotal=P0(1−α)+P0α+P0α=P0(1+α)P_{\text{total}} = P_0(1 - \alpha) + P_0\alpha + P_0\alpha = P_0(1 + \alpha)
Given: Ptotal=1P_{\text{total}} = 1 atm
P0(1+α)=1⇒P0=11+αP_0(1 + \alpha) = 1 \Rightarrow P_0 = \frac{1}{1 + \alpha}

3. Equilibrium Constant Expression KpK_p​

Kp=PCO⋅PCl2PCOCl2=(P0α)2P0(1−α)=P0α21−αK_p = \frac{P_{\text{CO}} \cdot P_{\text{Cl}_2}}{P_{\text{COCl}_2}} = \frac{(P_0\alpha)^2}{P_0(1 - \alpha)} = \frac{P_0\alpha^2}{1 - \alpha}
Substitute P0=11+αP_0 = \frac{1}{1 + \alpha}:
Kp=α2(1−α)(1+α)=α21−α2K_p = \frac{\alpha^2}{(1 - \alpha)(1 + \alpha)} = \frac{\alpha^2}{1 - \alpha^2}
Now, assume α=0.32\alpha = 0.32 (based on options):
Kp=(0.32)21−(0.32)2=0.10241−0.1024=0.10240.8976≈0.1141K_p = \frac{(0.32)^2}{1 - (0.32)^2} = \frac{0.1024}{1 - 0.1024} = \frac{0.1024}{0.8976} \approx 0.1141

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Condition 2: 0.4 atm of inert gas added, total pressure = 1 atm​

Here, 0.4 atm of an inert gas is added, and total pressure remains 1 atm.
Let new initial pressure of COCl₂ be P0′P_0', and new degree of dissociation = α′\alpha'

Species	Initial Pressure	Change	Equilibrium Pressure
COCl₂(g)	P0′P_0'	−P0′α′-P_0'\alpha'	P0′(1−α′)P_0'(1 - \alpha')
CO(g)	0	+P0′α′+P_0'\alpha'	P0′α′P_0'\alpha'
Cl₂(g)	0	+P0′α′+P_0'\alpha'	P0′α′P_0'\alpha'
Inert Gas	0.4 atm	0	0.4 atm

1. Total Equilibrium Pressure:​

Ptotal=P0′(1+α′)+0.4=1⇒P0′(1+α′)=0.6⇒P0′=0.61+α′P_{\text{total}} = P_0'(1 + \alpha') + 0.4 = 1 \Rightarrow P_0'(1 + \alpha') = 0.6 \Rightarrow P_0' = \frac{0.6}{1 + \alpha'}

2. Use the same KpK_p (temperature constant):​

Kp=(P0′α′)2P0′(1−α′)=P0′α′21−α′K_p = \frac{(P_0'\alpha')^2}{P_0'(1 - \alpha')} = \frac{P_0'\alpha'^2}{1 - \alpha'}
Substitute P0′=0.61+α′P_0' = \frac{0.6}{1 + \alpha'}:
Kp=0.6α′2(1−α′)(1+α′)=0.6α′21−α′2K_p = \frac{0.6 \alpha'^2}{(1 - \alpha')(1 + \alpha')} = \frac{0.6 \alpha'^2}{1 - \alpha'^2}
Equating to earlier KpK_p:
0.1141=0.6α′21−α′2⇒0.1141(1−α′2)=0.6α′2⇒0.1141=0.7141α′2⇒α′2=0.11410.7141≈0.15978⇒α′≈0.15978≈0.3997≈0.40.1141 = \frac{0.6 \alpha'^2}{1 - \alpha'^2} \Rightarrow 0.1141 (1 - \alpha'^2) = 0.6 \alpha'^2 \Rightarrow 0.1141 = 0.7141 \alpha'^2 \Rightarrow \alpha'^2 = \frac{0.1141}{0.7141} \approx 0.15978 \Rightarrow \alpha' \approx \sqrt{0.15978} \approx 0.3997 \approx 0.4

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Final Answer:​

• 
  
• Under condition 1, α≈0.32\alpha \approx 0.32
• 
  
• Under condition 2, α′≈0.4\alpha' \approx 0.4

Correct Option: (B) 0.32 and 0.4

 

[patelrahul]

Find the attached solution, ask if any confusion

 

[Kailaskumaran Anand]

Find the attached solution, ask if any confusion

Could you explain how you got the value of Kp?

 

[Kailaskumaran Anand]

Can someone explain how we got the value of Kp?