Question

The equilibrium concentration of A, B and C for the reaction A B + C are 4.6, 2.3 and
2.3 mole/ Mt respectively at 25°C. If 2 moles per litre of A are removed calculate the equilibrium
concentration of A, B and C at the same temperature.​

Answer 1

Answer:

A=B+C

Here A=4.6 and B=2.3 and C=2.3

so K=$\frac{[B][C]}{[A]}$

K=$\frac{2.3*2.3}{4.62}$

K=1.15 moles

A   =  B  +  C

4.6 2.3     2.3

-2   -2       -2

2.6  0.3   0.3

Q=$\frac{0.3*0.3}{2.6}$

so Q<K

reaction will shift forward

A→   B    +   C

2.6 0.3      0.3

-x    +x        +x

2.6-x 0.3+x  0.3+x

1.15=$\frac{(0.3+x)(0.3+x)}{2.6-x}$

1.15(2.6-x)=0.09+$x^{2}$+0.6x

2.99-1.15x=0.09+$x^{2}$+0.6x

$x^{2}$+1.75x-2.90=0

x=1.04

[A]=1.56mol/lit

[B]=1.34mol/lit

[C]=1.34mol/lit