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.?

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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

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

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