For a zero-order reaction- A- B- a graph of rate vs time has a slope equal to- -where k-rate constant-

The correct option is C zerofor zero order reaction #160;rate = k[a]^0 For a zero order reaction, the rate does not change #160;with timeThus, the slop ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Zero Order Reaction

For a zero-or...

Question

For a zero-order reaction, $A \to B$ , a graph of rate vs time has a slope equal to:
[where k=rate constant]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

-k

No worries! We‘ve got your back. Try BYJU‘S free classes today!

zero

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

-2.303 k

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C zero
for zero-order reaction

rate = $k [a]^{0}$

For a zero-order reaction, the rate does not change with time

Thus, the slope will be equal to zero.

Hence, the correct option is $C$

Suggest Corrections

Similar questions

Q. Rate constant

k

of a reaction varies with temperature according to the equation

log k = c o n s t a n t - \frac{E_{a}}{2.303 R} \times \frac{1}{T}

where

E_{a}

is the energy of activation for the reaction. When a graph is plotted for

log k

\frac{1}{T}

a straight line with a slope

- 6670 k

is obtained. The activation energy for this reaction will be:

(

R = 8.314 J K^{- 1} {m o l}^{- 1})

Q. For a zero-order reaction with a rate constant k, the slope of the plot of reactant concentration against time is:

Q. If the rate constant for a first order reaction is k, the time (t) required for the completion of 99% of the reaction is given by:

Q. Consider the following statements:
(i) Increase in concentration of reactant increases the rate of a zero order reaction.
(ii) Rate constant

k

is equal to collision frequency

A

E_{a} = 0

.
(iii) Rate constant

k

is equal to collision frequency

A

E_{a} = \infty

.
(iv)

{log}_{e} k

T

is a straight line.
(v)

{log}_{e} k

1 / T

is a straight line.
Correct statements are:

Q. For a first order reaction, the plot of ‘t’ against log C gives a straight line with slope equal to:

Join BYJU'S Learning Program

For a zero-order reaction, A→B, a graph of rate vs time has a slope equal to: [where k=rate constant]

The correct option is C zerofor zero-order reaction rate =k[a]0For a zero-order reaction, the rate does not change with timeThus, the slope will be equal to zero.Hence, the correct option is C

For a zero-order reaction, $A \to B$ , a graph of rate vs time has a slope equal to:
[where k=rate constant]

The correct option is C zero
for zero-order reaction

rate = $k [a]^{0}$

For a zero-order reaction, the rate does not change with time

Thus, the slope will be equal to zero.

Hence, the correct option is $C$