Test: Temperature Dependence of the Rate of a Reaction (November 24) - NEET MCQ

# Test: Temperature Dependence of the Rate of a Reaction (November 24) - NEET MCQ

Test Description

## 10 Questions MCQ Test Daily Test for NEET Preparation - Test: Temperature Dependence of the Rate of a Reaction (November 24)

Test: Temperature Dependence of the Rate of a Reaction (November 24) for NEET 2024 is part of Daily Test for NEET Preparation preparation. The Test: Temperature Dependence of the Rate of a Reaction (November 24) questions and answers have been prepared according to the NEET exam syllabus.The Test: Temperature Dependence of the Rate of a Reaction (November 24) MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Temperature Dependence of the Rate of a Reaction (November 24) below.
Solutions of Test: Temperature Dependence of the Rate of a Reaction (November 24) questions in English are available as part of our Daily Test for NEET Preparation for NEET & Test: Temperature Dependence of the Rate of a Reaction (November 24) solutions in Hindi for Daily Test for NEET Preparation course. Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free. Attempt Test: Temperature Dependence of the Rate of a Reaction (November 24) | 10 questions in 20 minutes | Mock test for NEET preparation | Free important questions MCQ to study Daily Test for NEET Preparation for NEET Exam | Download free PDF with solutions
Test: Temperature Dependence of the Rate of a Reaction (November 24) - Question 1

### How many times the rate of reaction increases at 200C for a reaction having the activation energies in the presence and absence of catalyst as 50 kJ/mol and 75 kJ/mol?

Detailed Solution for Test: Temperature Dependence of the Rate of a Reaction (November 24) - Question 1

K = Ae^(-Ea/RT)

Taking ln on both sides

lnk = lnA + (-Ea/R) x 1/T

Now ATQ: lnk1 = lnA + (-Ea1/R) x 1/T

lnk2 = lnA + (-Ea2/R) x 1/T

Subtracting the two equations

ln(k2/k1) = (Ea1 - Ea2)/RT

ln(k2/k1) = (75 - 50) x 1000 / 8.314 x 293

K2/K1 = e (25000 / 8.314 x 293)

K2/K1 = e10.26

This is approximately equal to 30000.

Hence C is correct.

Test: Temperature Dependence of the Rate of a Reaction (November 24) - Question 2

### If Ef and Eb are the activation energies of the forward and reverse reactions and the reaction is known to be exothermic, then:

Detailed Solution for Test: Temperature Dependence of the Rate of a Reaction (November 24) - Question 2

The correct answer is Option B.

For exothermic reaction, ΔH<0
Eb = Ef + ∣ΔH∣ or
Ef < Eb

 1 Crore+ students have signed up on EduRev. Have you?
Test: Temperature Dependence of the Rate of a Reaction (November 24) - Question 3

### The ratio of the rate constant of a reaction at two temperatures differing by __________0C is called temperature coefficient of reaction.

Detailed Solution for Test: Temperature Dependence of the Rate of a Reaction (November 24) - Question 3

The ratio of the rate constant of a reaction at two temperatures differing by 100C is called temperature coefficient of reaction.

Test: Temperature Dependence of the Rate of a Reaction (November 24) - Question 4

The effect of temperature on reaction rate is given by

Detailed Solution for Test: Temperature Dependence of the Rate of a Reaction (November 24) - Question 4
Arrhenius Equation

• Definition: The Arrhenius equation describes the relationship between the rate constant of a reaction and the temperature at which the reaction occurs.

• Formula: The Arrhenius equation is given by: k = A * e^(-Ea/RT), where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the ideal gas constant, and T is the temperature in Kelvin.

• Effect of Temperature: As per the Arrhenius equation, an increase in temperature leads to an increase in the rate constant of the reaction. This is because higher temperatures provide the reactant molecules with more energy, allowing them to overcome the activation energy barrier more easily.

• Application: The Arrhenius equation is widely used in chemical kinetics to predict how changes in temperature will affect the rate of a reaction.

Test: Temperature Dependence of the Rate of a Reaction (November 24) - Question 5

The activation energies of two reactions are given as Ea1= 40 J and Ea2= 80 J, then the relation between their rate constants can be written as:

Detailed Solution for Test: Temperature Dependence of the Rate of a Reaction (November 24) - Question 5

The correct answer is Option A.

As the value of activation energy, Ea increases, the value of rate constant, k decreases.

So, k1 > k2 since E1 < E2

Test: Temperature Dependence of the Rate of a Reaction (November 24) - Question 6

The activation energy of a chemical reaction can be determined by

Detailed Solution for Test: Temperature Dependence of the Rate of a Reaction (November 24) - Question 6

The correct answer is Option C
Activation energy can be determined by evaluation rate constants at different temperatures using equation:

Where, K2 = rate constant at temperature T2
K1= rate constant at temperature T1

Test: Temperature Dependence of the Rate of a Reaction (November 24) - Question 7

The rate constant, activation energy and Arrhenius parameter of a chemical reaction at 25°C are 3.0 x 10-4 s-1, 104.4 kJ mol-1 and 6.0 x 1014 s-1 respectively. The value of the rate constant at infinite temperature is is

Detailed Solution for Test: Temperature Dependence of the Rate of a Reaction (November 24) - Question 7

The correct answer is option C
Arrhenius equation
⇒ K = Ae−Ea/RT
As T → ∞,
RT → ∞

e−Ea/RT  → 1
Hence K → A as T→∞
∴ Value of K as T → ∞ = 6.0×1014S−1

Test: Temperature Dependence of the Rate of a Reaction (November 24) - Question 8

The plot of ln k vs 1/T is a straight line. The slope of the graph is:

Detailed Solution for Test: Temperature Dependence of the Rate of a Reaction (November 24) - Question 8

The Arrhenius equation is k=Ae−Ea​/RT

ln k = ln A - Ea/RT

comparing with y = mx+c where y = ln k and x = 1/T, slope m becomes -Ea/R

so when ln k vs 1/T is plotted, the slope comes out to be - Ea/R

Test: Temperature Dependence of the Rate of a Reaction (November 24) - Question 9

For a chemical reaction the rate constant is nearly doubled with the rise in temperature by

Detailed Solution for Test: Temperature Dependence of the Rate of a Reaction (November 24) - Question 9

Rate constant is doubled with 10 degree rise in temperature.

Test: Temperature Dependence of the Rate of a Reaction (November 24) - Question 10

The reactions with low activation energy are

Detailed Solution for Test: Temperature Dependence of the Rate of a Reaction (November 24) - Question 10

The correct answer is Option C.

The reactions with low activation energies are always fast whereas the reactions with high activation energy are always slow. The process of speeding up a reaction by reducing its activation energy is known as catalysis, and the factor that's added to lower the activation energy is called a catalyst.

## Daily Test for NEET Preparation

12 docs|366 tests
Information about Test: Temperature Dependence of the Rate of a Reaction (November 24) Page
In this test you can find the Exam questions for Test: Temperature Dependence of the Rate of a Reaction (November 24) solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Temperature Dependence of the Rate of a Reaction (November 24), EduRev gives you an ample number of Online tests for practice

## Daily Test for NEET Preparation

12 docs|366 tests