Chemical Engineering - Chemical Reaction Engineering - Discussion
Discussion Forum : Chemical Reaction Engineering - Section 8 (Q.No. 5)
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 ?
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 ?Discussion:
4 comments Page 1 of 1.
Anju Singh said:
7 years ago
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.
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.
(1)
Sushmita kumar said:
5 years ago
Formula for equilibrium conversion vs equilibrium constant is: X eq = K/1+K, where K is equilibrium constant.
Sachin said:
7 years ago
Explain the Solution please.
(1)
Balemual kumelachew said:
8 years ago
Thank you.
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