The rates of a certain reaction (dc/dt) at di | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The rate of a reaction is defined as the change in concentration of reactants or products per unit time. For a zero-order reaction,the rate of rea

Vedclass · Accepted Answer

Zero order

Vedclass · Answer

(A) The rate of a reaction is defined as the change in concentration of reactants or products per unit time. For a zero-order reaction,the rate of reaction is independent of the concentration of the reactants,i.e.,$	ext{Rate} = k[A]^0 = k$. In the given data,the rate of the reaction remains approximately constant $(2.8 	imes 10^{-2} \ mole \ litre^{-1} \ sec^{-1})$ at all given time intervals $(0, 10, 20, 30 \ sec)$. Since the rate does not change with time,the reaction must be of zero order.

Time $(sec)$	Rate $(mole \ litre^{-1} \ sec^{-1})$
$0$	$2.8 \times 10^{-2}$
$10$	$2.78 \times 10^{-2}$
$20$	$2.81 \times 10^{-2}$
$30$	$2.79 \times 10^{-2}$

The rates of a certain reaction $(dc/dt)$ at different times are as follows:
Time $(sec)$ Rate $(mole \ litre^{-1} \ sec^{-1})$
$0$ $2.8 \times 10^{-2}$
$10$ $2.78 \times 10^{-2}$
$20$ $2.81 \times 10^{-2}$
$30$ $2.79 \times 10^{-2}$

The reaction is:

Similar Questions

The unit of the velocity constant in case of zero order reaction is

For a zero order reaction,which of the following statements is false?

The rate constant for a zero order reaction is $3 \times 10^{-2} \ mol \ L^{-1} \ s^{-1}$. After $25 \ s$,if the concentration of the reactant is $0.5 \ M$,then the initial concentration of the reactant is ............ $M$.

What is the value of the slope of a graph obtained by plotting the concentration of reactant $[A]_t$ versus time for a zero order reaction?

For a zero-order reaction $A \rightarrow \text{Product}$,the half-life is $1 \ h$. If the initial concentration of reactant $A$ is $2.0 \ mol \ L^{-1}$,how many hours will it take for the concentration to decrease from $0.50 \ mol \ L^{-1}$ to $0.25 \ mol \ L^{-1}$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The rates of a certain reaction $(dc/dt)$ at different times are as follows:Time $(sec)$Rate $(mole \ litre^{-1} \ sec^{-1})$$0$$2.8 \times 10^{-2}$$10$$2.78 \times 10^{-2}$$20$$2.81 \times 10^{-2}$$30$$2.79 \times 10^{-2}$The reaction is: