Chemical Kinetics NEET 2024: Activation Energy Solution

What is the correct answer to the Chemical Kinetics NEET 2024 activation-energy question?

Option C, 18219 J as printed, is correct for this Chemical Kinetics NEET 2024 question. Using the supplied values gives approximately 18222.65 joules per mole, so C is the nearest option, not an exact match.

The question comes from Chemistry, Chemical Kinetics, in the NEET paper dated 23-06-2024. It asks for the activation energy of a reaction whose rate constant rises from 0.04 per second at 500 K to 0.14 per second at 700 K. The supplied constants are:

\log_{10} 3.5 = 0.5441, R = 8.31 J K^{- 1} m o l^{- 1}

This question bank classifies it as hard and gives an expected solving time of 120 seconds. That is the bank’s estimate, not a measured student average.

How do you pair the rate constants and temperatures correctly?

Pair each rate constant with its own temperature, then use the official solution’s two-temperature Arrhenius equation. Put the higher-temperature rate constant in the numerator and subtract the higher-temperature reciprocal from the lower-temperature reciprocal. This keeps both sides positive.

T_{1} = 500 K, k_{1} = 0.04 s^{- 1}

T_{2} = 700 K, k_{2} = 0.14 s^{- 1}

Use the following relation. Substitute without changing the ordering:

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

\ln (\frac{0.14}{0.04}) = \frac{E_{a}}{8.31} (\frac{1}{500} - \frac{1}{700})

The rate-constant ratio simplifies to 3.5. Identical units cancel, leaving a dimensionless ratio inside the logarithm:

\frac{0.14 s^{- 1}}{0.04 s^{- 1}} = 3.5

Since 3.5 exceeds one, its natural logarithm is positive; the reciprocal-temperature difference is also positive. A negative activation energy here means the ordering needs checking.

\ln 3.5 > 0, \frac{1}{500} - \frac{1}{700} > 0 \Rightarrow E_{a} > 0

How do you simplify the temperature term and convert log to ln?

Keep the reciprocal-temperature difference as an exact fraction, and convert the supplied common logarithm to a natural logarithm using the factor 2.303. The Arrhenius equation above requires the natural logarithm, so substituting the supplied common-log value directly would omit this factor.

First, carry out the reciprocal subtraction:

\frac{1}{500} - \frac{1}{700} = \frac{700 - 500}{500 \times 700} = \frac{200}{350000} = \frac{1}{1750} K^{- 1}

Keep this fraction rather than replacing it with a rounded decimal. It makes the rearrangement simpler and avoids another source of rounding.

Convert the logarithm, then substitute it into the equation:

\ln 3.5 = 2.303 \log_{10} 3.5 = 2.303 \times 0.5441

2.303 \times 0.5441 = \frac{E_{a}}{8.31} (\frac{1}{1750})

Multiply both sides by the gas constant and by 1750:

E_{a} = 0.5441 \times 2.303 \times 8.31 \times 1750

Keep all four supplied decimal places in 0.5441. Rounding it to 0.54 causes a separate precision error, demonstrated below.

Why is option C correct when the calculation gives 18222.65?

C is the closest option, although its printed value does not exactly equal the calculated result. Following the official worked solution gives this multiplication sequence: $0.5441 \times 2.303 = 1.2530623$ $1.2530623 \times 8.31 = 10.412947713$ $10.412947713 \times 1750 = 18222.65849775 \approx 18222.65$

This gives, consistent with the official solution:

E_{a} \approx 1.822 \times 10^{4} J m o l^{- 1} = 18.22 k J m o l^{- 1}

Compare the calculated number with each printed option:

C, 18219 J: about 3.65 J away.
D, 18030 J: about 192.65 J away.
B, 18500 J: about 277.35 J away.
A, 182310 J: an order of magnitude larger.

The supplied molar gas constant produces activation energy in joules per mole, although the options abbreviate the unit to J. Select option C, 18219 J as printed. The supplied calculation establishes the nearest answer, but does not explain how the slightly different option value was generated.

How can premature rounding change the answer from C to D?

Rounding the supplied common logarithm to 0.54 before multiplication makes D the nearest option, while leaving the Arrhenius equation unchanged. The following error path isolates premature rounding as the only change.

Error path: discard two supplied decimal places first.

\log_{10} 3.5 = 0.5441 \overset{premature rounding}{\to} 0.54

Keep every other factor unchanged:

E_{a} = 0.54 \times 2.303 \times 8.31 \times 1750 \approx 18085.34 J m o l^{- 1}

Compare distances from D and C: $| 18085.34 - 18030 | = 55.34$ $| 18085.34 - 18219 | = 133.66$

The rounded calculation selects D. The mistake is discarding supplied precision before comparing closely spaced options, not choosing the wrong equation.

This demonstrates a route to a distractor; it does not show that students commonly choose D. Preserve 0.5441 through the multiplication and round only the final product.

If this error repeats, record it separately from formula mistakes using NEET mock test error analysis. Name the exact step where precision was lost.

Which three follow-up questions check the same Chemical Kinetics method?

Use the same data to check temperature ordering, logarithmic slope and first-order half-life. These are original follow-up questions, not additional verified PYQs, with a worked answer immediately after each question.

If you reverse the temperature labels, does activation energy change?

No. Activation energy stays unchanged if you reverse both temperature labels and their paired rate constants consistently. The logarithm and reciprocal-temperature difference both change sign:

\ln (\frac{0.04}{0.14}) = \frac{E_{a}}{R} (\frac{1}{700} - \frac{1}{500})

- \ln 3.5 = \frac{E_{a}}{R} (- \frac{1}{1750})

The negative signs cancel, leaving the same result. Reversing only the logarithm or only the temperature difference would introduce a sign error.

E_{a} \approx 18.22 k J m o l^{- 1}

What is the slope of common log of the rate constant against reciprocal temperature?

The slope is negative, with a value given below. The rate constant increases while reciprocal temperature decreases, so the numerator is positive and the denominator is negative.

m = \frac{\log_{10} (0.14) - \log_{10} (0.04)}{1 / 700 - 1 / 500} = \frac{0.5441}{1 / 700 - 1 / 500} = \frac{0.5441}{- 1 / 1750} = - 0.5441 \times 1750 = - 952.175 K

For this common-log plot, the slope includes the conversion factor 2.303. A natural-log plot does not:

m_{\log_{10} k} = - \frac{E_{a}}{2.303 R}

m_{\ln k} = - \frac{E_{a}}{R}

If the reaction is first order, what are its half-lives?

The half-lives are 17.325 seconds at 500 K and 4.95 seconds at 700 K. Use the first-order half-life relation:

\ln 2 \approx 0.693, t_{1 / 2} = \frac{0.693}{k}

t_{1 / 2} (500 K) = \frac{0.693}{0.04} = 17.325 s

t_{1 / 2} (700 K) = \frac{0.693}{0.14} = 4.95 s

The larger rate constant gives the shorter half-life. Before selecting an option, check the logarithm base, paired temperatures, units and retained decimal places, then compare distances from the final result.

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).

Frequently asked questions

What is the correct answer to the NEET 2024 activation energy question?

For the question in the NEET paper dated 23-06-2024, option C, 18219 J as printed, is correct. The supplied values give approximately 18.22 kJ/mol, making C the nearest option rather than an exact match. Using the supplied molar gas constant gives units of J/mol, although the options print J.

Which Arrhenius formula should I use for two temperatures?

Use ln(k2/k1) = (Ea/R)(1/T1 - 1/T2). For this question, pair k1 = 0.04 per second with T1 = 500 K and k2 = 0.14 per second with T2 = 700 K. Both the logarithm and the reciprocal-temperature difference are positive with this ordering.

Why is 2.303 used in the activation energy calculation?

The supplied value, log10(3.5) = 0.5441, is a common logarithm, while the natural-log Arrhenius equation requires ln(3.5). Convert it using ln(3.5) = 2.303 × 0.5441. Substituting 0.5441 directly into the natural-log equation omits the conversion factor.

Why do I get option D instead of C in this question?

Rounding the supplied logarithm from 0.5441 to 0.54 gives about 18085.34 J/mol, making D the nearest option. Keeping 0.5441 gives about 18.22 kJ/mol, which is closest to C. Retain all supplied decimal places through multiplication and round only the final result.