New chat PYQ Library Doubts Rank report

Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →

Chemical Kinetics and Nuclear Chemistry question

2020 · 4 Sep · Shift 1 · Q9

Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.

PYQ Library
JEE Main
Chemistry
Chemical Kinetics and Nuclear Chemistry
2020 · 4 Sep · Shift 1 · Q9

Chemical Kinetics and Nuclear Chemistry question

2020 · 4 Sep · Shift 1 · Q9

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1

If 75% of a first order reaction was completed in 90 minutes, 60% of the same reaction would be completed in approximately (in minutes)

\underline{\hspace{2cm}}

. (Take : log 2 = 0.30; log 2.5 = 0.40)

Numerical answer

View written solutionFree

Correct answer: 60

Use the first-order integrated rate law

For a first-order reaction,

kt = 2.303 \log \frac{a}{a-x}

where:

$a$ = initial amount
$x$ = amount reacted after time $t$

So the fraction unreacted is $\dfrac{a-x}{a}$ .

Given: 75% reaction completed in 90 min

If $75\%$ is completed, then $25\%$ remains. Thus,

\frac{a}{a-x} = \frac{100}{25} = 4

So,

k(90) = 2.303 \log 4

Now,

\log 4 = \log(2^2) = 2\log 2 = 2(0.30)=0.60

Hence,

90k = 2.303(0.60)

Find time for 60% completion

If $60\%$ is completed, then $40\%$ remains. Thus,

\frac{a}{a-x} = \frac{100}{40} = 2.5

Let the required time be $t$ . Then,

kt = 2.303 \log 2.5

Given,

\log 2.5 = 0.40

So,

kt = 2.303(0.40)

Take ratio to eliminate $k$

From the two equations,

\frac{t}{90} = \frac{2.303(0.40)}{2.303(0.60)} = \frac{0.40}{0.60} = \frac{2}{3}

Therefore,

t = 90 \times \frac{2}{3} = 60 \text{ min}

Final answer

The time required for $60\%$ completion is

\boxed{60}

minutes.

More from Chemical Kinetics and Nuclear Chemistry

The number of molecules with energy greater than the threshold energy for a reaction increases five fold by a rise of temperature from 27oC to 42oC. Its energy of activation in J/mol is . (Take ln 5 = 1.6094; R =…2020 · Numerical
A flask contains a mixture of compounds A and B. Both compounds decompose by first-order kinetics. The half-lives for A and B are 300 s and 180 s, respectively. If the concentrations of A and B are equal initially, the time required for…2020 · MCQ
The rate constant (k) of a reaction is measured at differenct temperatures (T), and the data are plotted in the given figure. The activation energy of the reaction in kJ mol–1 is : (R is gas constant) Includes diagram2020 · MCQ
Consider the following reactions A $\to$ $P_{1}$ ; B $\to$ $P_{2}$ ; C $\to$ $P_{3}$ ; D $\to$ $P_{4}$ , The order of the above reactions are a, b, c, and d, respectively. The following graph is obtained when log[rate] vs. log[conc.] are plotted… Includes diagram2020 · MCQ
The rate of a reaction decreased by 3.555 times when the temperature was changed from 40oC to 30oC. The activation energy (in kJ mol–1) of the reaction is . Take; R = 8.314 J mol–1 K–1 ln 3.555 = 1.2682020 · Numerical
During the nuclear explosion, one of the products is 90Sr with half life of 6.93 years. If 1 $μ$ g of 90Sr was absorbed in the bones of newly born baby in placed of Ca, how much time, in years, is required to reduce much time, in year,…2020 · Numerical
For the reaction 2 $H_{2}$ (g) + 2 $N O$ (g) $\to$ $N_{2}$ (g) + 2 $H_{2} O$ (g) the observed rate expression is, rate = Kf[ $N O$ ]2[ $H_{2}$ ]. The rate expression for the reverse reaction is :2020 · MCQ
The rate of a certain biochemical reaction at physiological temperature (T) occurs 106 times faster with enzyme than without. The change in the activation energy upon adding enzyme is :2020 · MCQ