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For the reaction A + 3B = 2C + D (reversible), initial mole of A is twice that of B.if at equlibrium moles of B and C are equal .then the percent of B reacted is.?

For the reaction A + 3B = 2C + D (reversible), initial mole of A is twice that of B.if at equlibrium moles of B and C are equal .then the percent of B reacted is.?

Grade:12

3 Answers

Naveen Kumar
askIITians Faculty 60 Points
9 years ago
Given the reaction is:
.....................A......... + …......3B …...= …..2C..... +...... D
......................a......................a/2................0..............0....................(initially)
......................a-x....................(a/2)-3x..........2x.............x................(at equilibrium)
Here according the equation, 1 mole of A will combine with 3 mole of B and so if x mole of A is consumed then it will combine with 3x mole of B and similarly we can calculate the number of mole of C and D formed(according to their stoichiometric coefficient: just think about this stoichiometry : its very basic and important!!!!)
Now from the question given that, at equilibrium,
(a/2)-3x = 2x
a/2= 5x,
x=a/10
so amount of B reacted=3x=3a/10
Initial amount of B=a/2
so % of B reacted= {3a/10}/{a/2} *100
=6/10 *100=60%
Shriya Mehrotra
12 Points
9 years ago
thanku sir
Shreya
13 Points
4 years ago
                        A        +        3B      =      2C      +      D 
Initial :               x                   X/2.              0             0
At equilibrium :  x-y.             X/2-3y.        2y.             Y
Now , it is given that at equilibrium moles of B and C are equal . Therefore, x/2-3y = 2y (from above equation at equilibrium)
So, x/2 = 5y and x = 10y and y = x/10
Initially y was x/2 .. the final y has become x/10 
Therefore % of B reacted becomes final/initial *100
hence , (x/2 divided by x/10 )*100 = 20% 
 

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