Question

2A=3B+4C+D above reaction is taking place in a close container of volume 1 litre if initial no. Of moles of A is 10 which is .625 after 120 minutes then number of moles of B after 30 min
( Consider first order reaction)

Answer 1

Answer: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.?

Explanation:

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%

Answer 2

Given:

Equation of reaction2A=3B+4C+D

moles of A is 10 which is .625 after 120 minutes then number of moles of B after 30 min

Explanation:

Given the reaction is:

2A = 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.

Now from the given question, 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 the % of B reacted

= {3a/10}/{a/2} ×100

=6/10 ×100

=60%

Answer = 60%