back to directory
NEET CHEMISTRYChemical KineticsMedium

Question

The given graph is a representation of the kinetics of a reaction. The yy and xx axes for zero and first-order reactions, respectively are:

A

zero order (y=ratey = \text{rate} and x=concentrationx = \text{concentration}), first order (y=ratey = \text{rate} and x=t1/2x = t_{1/2})

B

zero order (y=concentrationy = \text{concentration} and x=timex = \text{time}), first order (y=t1/2y = t_{1/2} and x=concentrationx = \text{concentration})

C

zero order (y=concentrationy = \text{concentration} and x=timex = \text{time}), first order (y=rate constanty = \text{rate constant} and x=concentrationx = \text{concentration})

D

zero order (y=ratey = \text{rate} and x=concentrationx = \text{concentration}), first order (y=t1/2y = t_{1/2} and x=concentrationx = \text{concentration})

Step-by-Step Solution

Although the graph is not explicitly provided, the options imply a plot that yields a straight line parallel to the xx-axis (i.e., y=constanty = \text{constant}).

  1. For a zero-order reaction, the rate equation is Rate=k[R]0=k\text{Rate} = k[R]^0 = k. This means the rate is constant and independent of the concentration of the reactant. Thus, a plot of rate (yy-axis) versus concentration (xx-axis) is a straight horizontal line.
  2. For a first-order reaction, the half-life is given by t1/2=0.693kt_{1/2} = \frac{0.693}{k} . This means the half-life is a constant value and independent of the initial concentration of the reactant . Thus, a plot of t1/2t_{1/2} (yy-axis) versus concentration (xx-axis) is also a straight horizontal line. Therefore, the correct axes corresponding to a constant yy-value are: zero order (y=ratey = \text{rate} and x=concentrationx = \text{concentration}), first order (y=t1/2y = t_{1/2} and x=concentrationx = \text{concentration}).

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 Kineticsrepresentationkineticsreactionfirstorderreactions

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

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

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 →