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when 3 mole of A and 1 mole of B are mixed in a 1 lit vessel,the following reaction takes place A(g) + B(g)= 2C(g).reversible)..the euilibrium mixture contains 0.5 mole of C .what is the value of equil constant for the reaction.?

when 3 mole of A and 1 mole of B are mixed in a 1 lit vessel,the following reaction takes place A(g) + B(g)= 2C(g).reversible)..the euilibrium mixture contains 0.5 mole of C .what is the value of equil constant for the reaction.?

Grade:12

3 Answers

Naveen Kumar
askIITians Faculty 60 Points
9 years ago
Hello student!!
please follow the answers and try to understand it properly, if still having some doubts then don’t hesitate in posting on the forum.

in the given question, volume=1L, so initial concentration of A=3mol/1L=3M
Similarly for B=1M
Now,
A(g) + B(g)= 2C(g)
3mol/L.....1mol/L..0.......(initially)
3-x..........1-x …......2x...at equilibrium
But given that, mole of C=0.5 at equilibrium, Hence its equilibrium concentration=0.5/1L=0.5M=2x(from our equation)
So x=0.25M
writing for the equilibrium constant
K=[C]2/[A]*[B]=(2x)2/(3-x)*(1-x)
Put here the value of x and calculate K
Note: we should always put the concentratin with their unit
Shriya Mehrotra
12 Points
9 years ago
thanku so much sir. i got the ans.
Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the solution to your problem.

in the given question,
volume=1L,
so initial concentration of A
=3mol/1L
=3M
Similarly for B=1M
Now, A(g) + B(g)= 2C(g) 3mol/L.....1mol/L..0.......
(initially) 3-x..........1-x …......2x...
at equilibrium But given that, mole of C=0.5 at equilibrium,
Hence its equilibrium concentration
=0.5/1L
=0.5M
=2x(from our equation)
So x=0.25M writing for the equilibrium constant
K=[C]2/[A]*[B]
=(2x)2/(3-x)*(1-x)
Put here the value of x and calculate K
Note: we should always put the concentratin with their unit

Thanks and Regards

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