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Chemical Kinetics and Nuclear Chemistry question

2006 · Shift 0 · Q34

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Chemical Kinetics and Nuclear Chemistry
2006 · Shift 0 · Q34

Chemical Kinetics and Nuclear Chemistry question

2006 · Shift 0 · Q34

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1

A reaction was found to be second order with respect to the concentration of carbon monoxide. If the concentration of carbon monoxide is doubled, with everything else kept the same, the rate of reaction will

A
remain unchanged
B
triple
C
increase by a factor of 4
D
double

View written solutionFree

Correct answer: C

Use the meaning of reaction order

If a reaction is second order with respect to carbon monoxide, then the rate depends on carbon monoxide concentration as:

$\text{Rate} \propto [\mathrm{CO}]^2$

So we can write:

$\text{Rate} = k[\mathrm{CO}]^2$

where $k$ is the rate constant and all other quantities are unchanged.

Double the concentration of carbon monoxide

If the concentration of carbon monoxide is doubled, then:

$[\mathrm{CO}] \to 2[\mathrm{CO}]$

The new rate becomes:

$\text{New Rate} = k(2[\mathrm{CO}])^2$

Simplify

$\text{New Rate} = k \cdot 4[\mathrm{CO}]^2 = 4\,k[\mathrm{CO}]^2$

But

$k[\mathrm{CO}]^2 = \text{Original Rate}$

Therefore,

$\text{New Rate} = 4 \times \text{Original Rate}$

Evaluate the options

A: remain unchanged — Incorrect
B: triple — Incorrect
C: increase by a factor of 4 — Correct
D: double — Incorrect

Hence, the correct option is C.

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