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NEET CHEMISTRYChemical KineticsEasy

Question

Activation energy of any chemical reaction can be calculated if one knows the value of:

A

Probability of collision.

B

Orientation of reactant molecules during collision.

C

Rate constant at two different temperatures.

D

Rate constant at standard temperature.

Step-by-Step Solution

According to the Arrhenius equation, the relationship between the rate constant and temperature is given by: logk2k1=Ea2.303R[T2T1T1T2]\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right] where k1k_1 and k2k_2 are the rate constants at temperatures T1T_1 and T2T_2 respectively, EaE_a is the activation energy, and RR is the gas constant. Thus, if the rate constants at two different temperatures are known, the activation energy of the reaction can be easily calculated.

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 Kineticsactivationenergychemicalreactioncalculated

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

A.Internal energy
B.Enthalpy
C.Activation energy
D.Entropy
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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
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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
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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 $\frac{1}{T}$
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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.$8.3 \times 10^4$
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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}
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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 $10 \times 10^{–2} \text{ mol L}^{–1}$
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