Question

Determine the value of the equilibrium constant for the reaction.
$A+B\to 2C$
if 1.0 mol of $A$, 1.4 mol of $B$and 0.50 mol of $C$ are placed in a 1.0-L vessel and allowed come to equilibrium. The final concentration of $C$ is 0.75 mol/L.

Answer 1

$A+B⟶2C$

$t=0$ $1$ $1.4$ $0$

$t=t$ $1-x$ $1.4-x$ $2x$

given at equilibrium,

$2x=0.5$ mol

$x=0.25$ mol

at, $t=t$ $0.75$ $1.15$ $0.5$

So,

$K=\frac{{\left(0.5\right)}^2}{\left(0.75\right)\left(1.15\right)}$

$K=0.289$

Now, $2x=0.75$ mol

$x=0.375$ mol

So, $K=\frac{{\left(0.75\right)}^2}{(1-0.375)(1.4-0.375)}=0.878$

The final concentration of $C$ is $0.878$.