Question

Instantaneous rate of reaction for the reaction $3A+2B\to 5C$ is _____________
(a)- $+\dfrac{1}{3}\dfrac{d[A]}{dt}=-\dfrac{1}{2}\dfrac{d[B]}{dt}=+\dfrac{1}{5}\dfrac{d[C]}{dt}$
(b)- $-\dfrac{1}{3}\dfrac{d[A]}{dt}=-\dfrac{1}{2}\dfrac{d[B]}{dt}=+\dfrac{1}{5}\dfrac{d[C]}{dt}$
(c)- $-\dfrac{1}{3}\dfrac{d[A]}{dt}=+\dfrac{1}{2}\dfrac{d[B]}{dt}=-\dfrac{1}{5}\dfrac{d[C]}{dt}$
(d)- $+\dfrac{1}{3}\dfrac{d[A]}{dt}=-\dfrac{1}{2}\dfrac{d[B]}{dt}=-\dfrac{1}{5}\dfrac{d[C]}{dt}$

Answer 1

Hint: In the reaction of instantaneous rate of the reactants are written with the negative sign because in the reaction the reactants are decreasing and the products are written with the positive sign because in the reaction the products are increasing.

Complete step by step answer:
Let us see how to calculate the rate of reaction:
Suppose a reaction is
$R\to P$
So, the rate of reaction is equal to the decrease in the concentration of the R divided by the time interval or it is equal to the increase in the concentration of the P divided by the time interval. So, it is written as:
$\text{Rate of reaction = }\dfrac{\text{Decrease in concentration of R}}{\text{Time interval}}\text{ = }\dfrac{\text{Increase in concentration of P}}{\text{Time interval}}$
So, we can say that a change in concentration of any one of the reactants or products per unit time is the rate of reaction.
$\text{Rate of reaction = -}\dfrac{\Delta [R]}{\Delta t}=+\dfrac{\Delta [P]}{\Delta t}$

The negative sign of reactant implies that the concentration of the reactant is decreasing and the positive sign of the product implies that the concentration of the product is increasing.
When the rate of reaction is calculated at any instant of time, then it is called the instantaneous rate of reaction. The rate of reaction at any instant time is given by:
$\dfrac{dx}{dt}$.

So, the reaction given is:
$3A+2B\to 5C$
The rate of reaction for this equation will be:
$\text{Rate of reaction = -}\dfrac{1}{3}\dfrac{\Delta [A]}{\Delta t}=\text{-}\dfrac{1}{2}\dfrac{\Delta [B]}{\Delta t}=+\dfrac{1}{5}\dfrac{\Delta [C]}{\Delta t}$
Thus the instantaneous rate of reaction for this reaction will be:
 $\text{Instantaneous rate (}{{\text{r}}_{ins}})\text{ = }-\dfrac{1}{3}\dfrac{d[A]}{dt}=-\dfrac{1}{2}\dfrac{d[B]}{dt}=+\dfrac{1}{5}\dfrac{d[C]}{dt}$

So, the correct answer is $-\dfrac{1}{3}\dfrac{d[A]}{dt}=-\dfrac{1}{2}\dfrac{d[B]}{dt}=+\dfrac{1}{5}\dfrac{d[C]}{dt}$
So, the correct answer is “Option B”.

Note: The unit of concentration is expressed in$moles/liter$, and the time is expressed in seconds or minutes. So the unit of rate of reaction is usually expressed in $moles\ lite{{r}^{-1}}\text{ }{{\sec }^{-1}}$ or $moles\ lite{{r}^{-1}}\text{ mi}{{\text{n}}^{-1}}$ .