Question

The rate law for the reaction : $Ester+H^{+}\to Acid+Alcohol$ is
$V=k\left[ester\right]{\left[H_3{O}^{+}\right]}^0$
What would be the new rate if
(a)conc. of ester is doubled
(b)conc. of $H^{+}$ is doubled

• A. (a)$v$ (b)$2v$
• B. (a)$2v$ (b)$v$
• C. (a)$2v$ (b)$2v$
• D. None of the above

Answer 1

The correct option is B (a)$2v$ (b)$v$

$\upsilon =k\left[ester\right]\left[H_3{O}^{+}\right]$

(a) Conc. of ester is doubled rate also double that is $2\upsilon$ because rate of the reaction depends upon ester concentration.

(b) Conc. of $H^{+}$ is doubled rate does not change that is $\upsilon$ because rate of the reaction does not depends on $H_3{O}^{+}$ concentration.

So answer is B.