Chemical Engineering - Chemical Reaction Engineering - Discussion

5. 

For the reversible reaction A 2B, if the equilibrium constant K is 0.05 mole/litre; starting from initially 2 moles of A and zero moles of B, how many moles will be formed at equilibrium ?

[A]. 0.253
[B]. 0.338
[C]. 0.152
[D]. 0.637

Answer: Option B

Explanation:

No answer description available for this question.

Balemual Kumelachew said: (Oct 13, 2017)  
Thank you.

Sachin said: (Aug 3, 2018)  
Explain the Solution please.

Anju Singh said: (Feb 28, 2019)  
k = [product]^n/[reactant].

A = 2B
2 ----------- 0
2-x ----------- 2x

And total no of moles = 2x+2-x = 2+x.

K=(2x/2+x)^2/[(2-x)/(2+x)] put the value of k solved the equation.
x=.2238.

No of moles formed = 2* x = .4476.
So, option (B) is correct.

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