For an equation A + 3B → 4C + 2D; if −\( \displaystyle \frac{d[\mathrm{A}]}{dt} \) at time t, is 1.5 × 10−2 M s−1, the value of +\( \displaystyle \frac{d[\mathrm{C}]}{dt} \) at t will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 6.0 × 10−2 M s−1
The rate of a reaction must have one value, whichever species is used to measure it. So the rate of change of each concentration is divided by its stoichiometric coefficient, with a minus sign for reactants (they are used up) and a plus sign for products (they are formed):
\( \displaystyle \ce{A + 3B -> 4C + 2D} \)
\( \displaystyle \text{Rate}=-\frac{d[\mathrm{A}]}{dt}=-\frac{1}{3}\frac{d[\mathrm{B}]}{dt}=+\frac{1}{4}\frac{d[\mathrm{C}]}{dt}=+\frac{1}{2}\frac{d[\mathrm{D}]}{dt} \)
Relating A and C:
\( \displaystyle +\frac{d[\mathrm{C}]}{dt}=4\times\left(-\frac{d[\mathrm{A}]}{dt}\right) \)
\( \displaystyle +\frac{d[\mathrm{C}]}{dt}=4\times1.5\times10^{-2} \)
\( \displaystyle +\frac{d[\mathrm{C}]}{dt}=6.0\times10^{-2}\ \mathrm{M\,s^{-1}} \)
So C is formed four times as fast as A is used up: 6.0 × 10−2 M s−1.