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NEET CHEMISTRYChemical KineticsEasy

Question

The incorrect statement among the following is:

A

The value of the equilibrium constant is changed in the presence of a catalyst

B

Enzymes catalyse mainly bio-chemical reactions

C

Coenzymes increase the catalytic activity of the enzyme

D

The catalyst does not initiate any reaction

Step-by-Step Solution

A catalyst increases the rate of the chemical reaction by making available a new low energy pathway for the conversion of reactants to products. It increases the rate of forward and reverse reactions that pass through the same transition state and does not affect the equilibrium composition of a reaction mixture. Therefore, the value of the equilibrium constant is not changed in the presence of a catalyst . Enzymes are biological catalysts that catalyse biochemical reactions . Non-protein constituents called co-factors (which include coenzymes) bind to the enzyme to make it catalytically active . A catalyst generally does not initiate a reaction but only alters the rate of a spontaneous 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 Kineticsincorrectstatementfollowing

More Chemical Kinetics Questions

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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

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

A.Internal energy
B.Enthalpy
C.Activation energy
D.Entropy
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

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