Question

The rate law of gaseous reaction : $A_{\left(g\right)}+B\left(g\right)\to$ product is given by $K\left[A{]}^2\right[B]$. If the 'V' of one reaction vessel is suddenly doubled, which of the following will happen?

• A. The rate $w.r.t^{\prime }{A}^{\prime }$ will decrease Two times.
• B. The rate $w.r.t^{\prime }{A}^{\prime }$ will decrease Four times.
• C. The rate $w.r.t^{\prime }{B}^{\prime }$ will decrease Two times.
• D. The overall rate will decrease Eight Times of the original Value.

Answer 1

The correct option is C The rate $w.r.t^{\prime }{B}^{\prime }$ will decrease Two times.

Solution:-

$A_{\left(g\right)}+B_{\left(g\right)}⟶\text{Product}$

Rate constant for the above reaction-

$K={\left[A\right]}^2\left[B\right]$

Initial volume of the vessel is $V$.

$\therefore K=V^2.V=V^3$

Now if the volume of $1$ vessel is doubled, i.e., new volume of the vessel will be $2V$.

Active mass $=$ concentration in mole pre litre.

If the volume of $A$ vessel is doubled, the molar concentration of $A$ will decrease to its half.

$\therefore K^{\prime }={\left(\frac{V}{2}\right)}^2.V=\frac{V^3}{4}=\frac{K}{4}(\because V^3=K)$

Hence, if we double the volume of vessel $A$, the rate w.r.t. $A$ will decrease by $4$ times.

Similarly, if the volume of $B$ vessel is doubled, the molar concentration of $B$ will decrease to its half.

$\therefore K^{\prime \prime }=V^2.\frac{V}{2}=\frac{V^3}{2}=\frac{K}{2}$

Hence, if we double the volume of vessel $B$, the rate w.r.t. $B$ will decrease by $2$ times.