Question

If $A+B\rightleftharpoons C+D$; $K_C=K_1$ and $E^{\ominus }=aV$; $2A+2B\rightleftharpoons 2C+2D$; $K_C=K_2$ and $E^{\ominus }=bV$ then:

• A. $a=b$
• B. $K_2=K_1^2$
• C. $a=2b$
• D. $b=a^2$

Answer 1

Answer: (A), (B)

The correct options are
A $a=b$
B $K_2=K_1^2$

$K_1=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]}$ and $K_2=\frac{\left[C{]}^2\right[D{]}^2}{\left[A{]}^2\right[B{]}^2}$

Also, $E^{\ominus }$ is independent of stoichiometry.

Hence, $a=b$

Also, $K_2=K_1^2$ (from the expression given above)

Hence, options A and B are correct.