Chemical Kinetics NEET 2013: Rate-Doubling Solution

What is the answer to the Chemical Kinetics NEET 2013 rate-doubling question?

Option B, 34.7 kJ mol⁻¹, is correct for the Chemical Kinetics NEET 2013 rate-doubling question. Use absolute temperatures and keep the logarithm base consistent with the Arrhenius formula.

A reaction proceeds twice as fast at 35°C as at 20°C. Calculate its activation energy using the supplied gas constant:

R = 8.314 J m o l^{- 1} K^{- 1}

The supplied options are:

Source: NEET 2013 Chemistry, Chemical Kinetics, Arrhenius Equation and Activation Energy. This question bank labels the question hard and assigns an expected solving time of 90 seconds, not a time based on measured student performance.

How do you set up the two-temperature Arrhenius equation?

Assign the higher temperature to the faster reaction and use a rate-constant ratio of two. The comparison treats other rate-controlling conditions, including reactant concentrations, as unchanged, so doubling the reaction rate doubles the rate constant.

The usual assumption is that the Arrhenius pre-exponential factor remains effectively constant over this temperature interval. The official method is:

\ln (\frac{k_{2}}{k_{1}}) = \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})

Convert both temperatures using the official solution’s convention: $T_{1} = 20 + 273 = 293 K$ $T_{2} = 35 + 273 = 308 K$

The rate information gives:

\frac{k_{2}}{k_{1}} = 2

A temperature increase of 15°C is also an increase of 15 K. Reciprocal temperatures still require kelvin values. The equation uses absolute temperatures, not merely their difference.

Rearrange before inserting numbers to keep the numerator and denominator clear:

E_{a} = \frac{R \ln (k_{2} / k_{1})}{(\frac{1}{T_{1}} - \frac{1}{T_{2}})}

Substitute the given values:

E_{a} = \frac{8.314 \ln 2}{(\frac{1}{293} - \frac{1}{308})} J m o l^{- 1}

How does the calculation give option B?

The reciprocal-temperature difference is approximately 0.000166 inverse kelvin. Dividing the numerator by this small positive number gives an activation energy near 34,700 joules per mole. Convert that value to kilojoules per mole before matching the options.

First, combine the reciprocal terms:

\frac{1}{293} - \frac{1}{308} = \frac{308 - 293}{293 \times 308} = \frac{15}{90244} \approx 0.000166 K^{- 1}

Use the natural logarithm of two: $\ln 2 \approx 0.693$ $8.314 \times 0.693 \approx 5.762$

Following the supplied solution’s rounded calculation:

E_{a} \approx \frac{8.314 \times 0.693}{0.000166} J m o l^{- 1} \approx 34680 J m o l^{- 1} \approx 34.7 k J m o l^{- 1}

These intermediate values are approximations. Retaining more digits changes only the rounding, not the selected option.

Check the unit cancellation:

\frac{J m o l^{- 1} K^{- 1}}{K^{- 1}} = J m o l^{- 1}

Divide the numerical value in joules per mole by 1000 to obtain kilojoules per mole:

\frac{34680}{1000} = 34.68 \approx 34.7

Finally, check the sign. The second temperature is higher, so the reciprocal-temperature difference is positive:

T_{2} > T_{1} \Rightarrow \frac{1}{T_{1}} - \frac{1}{T_{2}} > 0

The natural logarithm of two is also positive, giving positive activation energy. The answer is option B, 34.7 kJ mol⁻¹.

How does a logarithm mismatch produce option C?

Inserting a common logarithm into the natural-log formula gives option C, 15.1 kJ mol⁻¹. This reproduces the option through a specific error; it does not establish how the examiner designed it.

The common-log value is: $\log_{10} 2 \approx 0.301$

The incorrect substitution is:

E_{a, w r o n g} = \frac{8.314 \times 0.301}{0.000166} \approx 15075 J m o l^{- 1} \approx 15.1 k J m o l^{- 1}

That result matches option C, but it uses the wrong logarithm base. Changing the base requires changing the coefficient: $\ln x \approx 2.303 \log_{10} x$

The correct common-log form, using the rounded conversion factor, is:

\log_{10} (\frac{k_{2}}{k_{1}}) = \frac{E_{a}}{2.303 R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})

When solving for activation energy, the missing factor belongs in the numerator:

E_{a} = \frac{2.303 R \log_{10} (k_{2} / k_{1})}{(\frac{1}{T_{1}} - \frac{1}{T_{2}})}

It restores the natural-log value: $2.303 \times 0.301 \approx 0.693$

The result returns to 34.7 kJ mol⁻¹. Pair natural log with the gas constant alone, or common log with 2.303 times the gas constant when solving for activation energy. Use the natural-log form here: it follows the official solution and avoids an extra conversion factor.

How do you solve two related Chemical Kinetics practice questions?

The answers are approximately 69.4 kJ mol⁻¹ for the fourfold-rate question and 33.3 kJ mol⁻¹ for the graph-slope question. Both are original chapter practice, not verified NEET PYQs. The first tests the effect of changing the rate ratio; the second uses the Arrhenius equation in straight-line form.

Original chapter practice 1: What if the rate becomes four times as large?

Under the same assumptions, a reaction’s rate becomes four times as large between 293 K and 308 K. Calculate its activation energy using:

R = 8.314 J m o l^{- 1} K^{- 1}

Keep the temperatures unchanged and replace the rate-constant ratio:

E_{a} = \frac{R \ln 4}{(\frac{1}{293} - \frac{1}{308})}

Use the logarithm identity: $\ln 4 = \ln (2^{2}) = 2 \ln 2$

The denominator is unchanged, while the numerator doubles:

E_{a} \approx 2 \times 34.7 = 69.4 k J m o l^{- 1}

Changing the rate ratio from two to four doubles the inferred activation energy for this fixed temperature interval because its logarithm doubles. It does not make activation energy four times as large.

Original chapter practice 2: How do you use an Arrhenius-plot slope?

A plot of natural log of the rate constant against reciprocal absolute temperature has the following slope. Calculate the activation energy using the same gas constant.

m = - 4.00 \times 10^{3} K

Start from the logarithmic Arrhenius equation:

\ln k = \ln A - \frac{E_{a}}{R T}

Comparing it with a straight-line equation gives:

m = - \frac{E_{a}}{R}

Therefore:

E_{a} = - m R = - (- 4.00 \times 10^{3}) (8.314) = 33256 J m o l^{- 1} \approx 33.3 k J m o l^{- 1}

The negative graph slope is consistent with positive activation energy. A common-log plot would instead have:

m = - \frac{E_{a}}{2.303 R}

Before substituting an Arrhenius-plot slope, underline the logarithm base on the vertical axis. That check tells you which coefficient to use.

Next step: the past-paper archive on NEET JEEnius AI and search past NEET papers by year, subject or chapter, each with a worked solution (100 free searches a month).

For a worked example of the same idea, see Doppler Effect NEET 2009: Car-and-Hill Echo at 720 Hz.

Frequently asked questions

What is the answer to the Chemical Kinetics NEET 2013 rate-doubling question?

Option B, 34.7 kJ/mol, is correct. With other rate-controlling conditions unchanged, use k₂/k₁ = 2, T₁ = 293 K and T₂ = 308 K in Eₐ = R ln(k₂/k₁)/(1/T₁ − 1/T₂). The result is approximately 34,700 J/mol, or 34.7 kJ/mol.

Why do we convert Celsius to kelvin in the Arrhenius equation?

The Arrhenius equation uses reciprocal absolute temperatures, so both temperatures must be in kelvin. Although a rise of 15°C equals a rise of 15 K, substituting 20 and 35 into the reciprocal terms is incorrect. Use 293 K and 308 K for this question.

Why am I getting 15.1 kJ/mol instead of 34.7 kJ/mol?

Using log₁₀2 ≈ 0.301 in the natural-log formula gives approximately 15.1 kJ/mol, matching option C. Either use ln 2 ≈ 0.693 with R or multiply the common-log numerator by 2.303. Both consistent methods give 34.7 kJ/mol.

How do I calculate activation energy from an Arrhenius plot?

For a plot of ln k against 1/T, the slope m equals −Eₐ/R, so Eₐ = −mR. A slope of −4.00 × 10³ K gives approximately 33.3 kJ/mol using R = 8.314 J mol⁻¹ K⁻¹. For a log₁₀ k plot, use Eₐ = −2.303mR instead.

Chemical Kinetics NEET 2013: Rate-Doubling Solution

What is the answer to the Chemical Kinetics NEET 2013 rate-doubling question?

How do you set up the two-temperature Arrhenius equation?

How does the calculation give option B?

How does a logarithm mismatch produce option C?

How do you solve two related Chemical Kinetics practice questions?

Frequently asked questions

What is the answer to the Chemical Kinetics NEET 2013 rate-doubling question?

Why do we convert Celsius to kelvin in the Arrhenius equation?

Why am I getting 15.1 kJ/mol instead of 34.7 kJ/mol?

How do I calculate activation energy from an Arrhenius plot?

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