For the equation 2A+BBA2 , the equilibrium concentration of A, B, BA2 is 4, 2 and 2 respectively. The value of Kc will be

Question

For the equation 2A+BBA2 , the equilibrium concentration of A, B, BA2 is 4, 2 and 2 respectively. The value of Kc will be (A)  0.0625  (B)  0.625  (C)  6.280  (D)  6.250

Tardigrade Admin · Accepted Answer

underset4 mathop2A + underset2 mathopB underset2 mathopBA2 Equilibrium text concentration Kc=([BA2]/[A]2[B]) =(2/(4)2× 2)=(1/16)=0.0625

Q. For the equation $2 A + BB A_{2}$ , the equilibrium concentration of $A, B, B A_{2}$ is 4, 2 and 2 respectively. The value of $K_{c}$ will be

$0.0625$

$0.625$

$6.280$

$6.250$

$4 2 A + 2 B 2 B A_{2}_{Eq u i l ib r i u m co n ce n t r a t i o n}$ $K_{c} = \frac{[ B A _{2} ]}{[ A ] ^{2} [ B ]}$ $= \frac{2}{( 4 ) ^{2} \times 2} = \frac{1}{16} = 0.0625$

Q. For the equation 2A+BBA2​ , the equilibrium concentration of A,B,BA2​ is 4, 2 and 2 respectively. The value of Kc​ will be

0.0625

0.625

6.280

6.250