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NEET CHEMISTRYChemical KineticsMedium

Question

The decomposition of phosphine (PH3\text{PH}_3) on tungsten at low pressure is a first-order reaction. It is because the:

A

rate is proportional to the surface coverage

B

rate is inversely proportional to the surface coverage

C

rate is independent of the surface coverage

D

rate of decomposition is very slow

Step-by-Step Solution

For a surface-catalyzed decomposition reaction, the rate depends on the fraction of the catalyst surface covered by the reactant molecules. At low pressures, the surface is sparsely covered, and the surface coverage is directly proportional to the partial pressure of the reactant gas. Since the rate is proportional to the surface coverage, it becomes proportional to the pressure of the gas, making it a first-order reaction. Conversely, at high pressures, the surface becomes fully saturated (completely covered), and the rate becomes independent of pressure, resulting in zero-order kinetics.

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 Kineticsdecompositionphosphinetextphtungstenpressure

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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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A.Internal energy
B.Enthalpy
C.Activation energy
D.Entropy
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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 $T$
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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.$4.0 \times 10^2$
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D.$8.3 \times 10^4$
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B.+\frac{d[NH_3]}{dt} = -\frac{2}{3}\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:

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