back to directory
NEET CHEMISTRYChemical KineticsEasy

Question

Which quantity is altered when a catalyst is introduced during a chemical reaction?

A

Internal energy

B

Enthalpy

C

Activation energy

D

Entropy

Step-by-Step Solution

A catalyst is a substance which increases the rate of a reaction without itself undergoing any permanent chemical change. It operates by providing an alternate pathway or reaction mechanism by reducing the activation energy between reactants and products, hence lowering the potential energy barrier. A catalyst does not alter the thermodynamic state functions such as internal energy, enthalpy, or entropy of the reaction.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Chemical Kinetics. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYChemical Kineticsquantityalteredcatalystintroducedduring

More Chemical Kinetics Questions

View all

If the half-life is independent of its initial concentration, then the order of the reaction is:

A.0
B.1
C.3
D.2
EasySolve

The rate of the reaction $2NO + Cl_2 \rightarrow 2NOCl$ is given by the rate equation $\text{rate} = k[NO]^2[Cl_2]$. The value of the rate constant can be increased by:

A.Increasing the concentration of $NO$
B.Increasing the concentration of $Cl_2$
C.Increasing the temperature
D.All of the above
EasySolve

The rate of a first-order reaction is $1.5 \times 10^{-2} \text{ mol L}^{-1} \text{ min}^{-1}$ at $0.5 \text{ M}$ concentration of the reactant. The half-life of the reaction is:

A.23.1 min
B.8.73 min
C.7.53 min
D.0.383 min
MediumSolve

The activation energy of a reaction can be determined from the slope of the graph between:

A.$\ln k$ vs $T$
B.$\frac{\ln k}{T}$ vs $T$
C.$\ln k$ vs $\frac{1}{T}$
D.$\frac{T}{\ln k}$ vs $\frac{1}{T}$
EasySolve

The plot of $\ln k$ vs $1/T$ for the following reaction, $2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)$ gives a straight line with the slope of the line equal to $-1.0 \times 10^4 \text{ K}$. What is the activation energy for the reaction in $\text{J mol}^{–1}$? (Given: $R = 8.3 \text{ J K}^{–1} \text{ mol}^{–1}$)

A.$4.0 \times 10^2$
B.$4.0 \times 10^{-2}$
C.$8.3 \times 10^{-4}$
D.$8.3 \times 10^4$
MediumSolve

Consider the reaction $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$. The equality relationship between $\frac{d[NH_3]}{dt}$ and $-\frac{d[H_2]}{dt}$ is:

A.\frac{d[NH_3]}{dt} = -\frac{1}{3}\frac{d[H_2]}{dt}
B.+\frac{d[NH_3]}{dt} = -\frac{2}{3}\frac{d[H_2]}{dt}
C.+\frac{d[NH_3]}{dt} = -\frac{3}{2}\frac{d[H_2]}{dt}
D.+\frac{d[NH_3]}{dt} = -\frac{d[H_2]}{dt}
EasySolve

For a chemical reaction, $4A + 3B \rightarrow 6C + 9D$ rate of formation of C is $6 \times 10^{–2} \text{ mol L}^{–1} \text{ s}^{–1}$ and rate of disappearance of A is $4 \times 10^{–2} \text{ mol L}^{–1} \text{ s}^{–1}$. The rate of reaction and amount of B consumed in interval of 10 seconds, respectively will be:

A.$1 \times 10^{–2} \text{ mol L}^{–1} \text{ s}^{–1}$ and $30 \times 10^{–2} \text{ mol L}^{–1}$
B.$10 \times 10^{–2} \text{ mol L}^{–1} \text{ s}^{–1}$ and $10 \times 10^{–2} \text{ mol L}^{–1}$
C.$1 \times 10^{–2} \text{ mol L}^{–1} \text{ s}^{–1}$ and $10 \times 10^{–2} \text{ mol L}^{–1}$
D.$10 \times 10^{–2} \text{ mol L}^{–1} \text{ s}^{–1}$ and $30 \times 10^{–2} \text{ mol L}^{–1}$
MediumSolve

The half-life period of a first-order reaction is 1386 s. The specific rate constant of the reaction is:

A.5.0 × 10⁻³ s⁻¹
B.0.5 × 10⁻² s⁻¹
C.0.5 × 10⁻³ s⁻¹
D.5.0 × 10⁻² s⁻¹
EasySolve

This neet chemistry practice question is part of the TopperSquare free question bank. TopperSquare offers 15,000+ chapter-wise NEET MCQs across Physics, Chemistry, and Biology with detailed step-by-step explanations, full mock tests, NEET PYQs (2010–2024), and an AI-powered performance analytics dashboard. browse all neet practice questions → · practice chemistry sets →