Arrhenius Equation NEET 2026: Activation Energy Solution

What is the answer to the Arrhenius Equation NEET 2026 question?

Option A, 24.84 kcal per mole, is correct for the Arrhenius Equation NEET 2026 question. The activation energy comes from the coefficient of reciprocal temperature, not the standalone constant 14.34.

For a first-order reaction, determine the activation energy in kcal per mole from:

\ln k = 14.34 - \frac{1.25 \times 10^{4}}{T}

Temperature is in kelvin. The rate constant has units of inverse seconds, and the supplied gas constant is:

[k] = s^{- 1}, R = 1.987 c a l m o l^{- 1} K^{- 1}

The supplied options, all in kcal per mole, are:

This Chemical Kinetics question is classified as medium by the question bank. Its expected solving time is 60 seconds, a question-bank target, not a measured student average.

How do you match the given equation to the Arrhenius form?

The coefficient of reciprocal temperature gives activation energy divided by the gas constant. The standalone constant gives the natural logarithm of the prefactor. Start with the Arrhenius equation:

k = A \exp (- \frac{E_{a}}{R T})

Take natural logarithms on both sides:

\ln k = \ln A + \ln [\exp (- \frac{E_{a}}{R T})] = \ln A - \frac{E_{a}}{R T}

Rewrite the temperature-dependent term as a coefficient multiplying reciprocal temperature:

\ln k = \ln A - (\frac{E_{a}}{R}) (\frac{1}{T})

Write the supplied expression in the same form:

\ln k = 14.34 - (1.25 \times 10^{4}) (\frac{1}{T})

Matching the constant and temperature-dependent terms separately gives: $\ln A = 14.34$

\frac{E_{a}}{R} = 1.25 \times 10^{4} K

The negative sign already belongs to the Arrhenius expression. It does not make the activation energy negative.

No numerical temperature is needed because the coefficient of reciprocal temperature is supplied. No factor of 2.303 is required: the equation uses natural logarithms, not common logarithms.

How do you calculate activation energy and convert cal to kcal?

Multiply the matched coefficient by the supplied gas constant, then divide by 1,000 to convert calories into kilocalories. The multiplication gives calories per mole, not the requested kilocalories per mole.

Rearrange the matched relationship:

E_{a} = (1.25 \times 10^{4} K) R

Substitute the given gas constant:

E_{a} = (1.25 \times 10^{4} K) (1.987 c a l m o l^{- 1} K^{- 1})

Kelvin cancels with inverse kelvin, leaving calories per mole:

K \times c a l m o l^{- 1} K^{- 1} = c a l m o l^{- 1}

The arithmetic is:

E_{a} = 12, 500 \times 1.987 = 24, 837.5 c a l m o l^{- 1}

Apply the unit conversion:

1 k c a l = 1, 000 c a l

E_{a} = \frac{24, 837.5}{1, 000} = 24.8375 k c a l m o l^{- 1}

Rounding gives:

E_{a} = 24.84 k c a l m o l^{- 1}

Select option A. For a magnitude check, approximate the gas constant by 2:

12, 500 \times 2 \approx 25, 000 c a l m o l^{- 1} = 25 k c a l m o l^{- 1}

First-order kinetics explains why the rate constant has units of inverse seconds. It does not introduce an extra factor into this calculation.

Why does choosing 14.34 give the wrong answer?

Choosing option B, 14.34, means taking the standalone constant and treating it as activation energy. That is a term-matching error: the constant represents the natural logarithm of the Arrhenius prefactor, while the coefficient attached to reciprocal temperature represents activation energy divided by the gas constant.

The correct matches are: $14.34 = \ln A$

1.25 \times 10^{4} K = \frac{E_{a}}{R}

A logarithmic intercept cannot be assigned energy units just because the question asks for kcal per mole. The energy units enter when the temperature coefficient is multiplied by the gas constant.

Rewrite both equations as constant minus coefficient times reciprocal temperature before matching terms. Keep the notation separate from the answer labels: the Arrhenius prefactor, written as capital A in the equation, is a physical parameter; option A labels the correct choice.

How do you solve related graph and common-logarithm questions?

For a straight-line graph, include the minus sign in the slope. For common logarithms, divide both terms by 2.303; the activation energy stays unchanged.

The following are related practice questions derived from the supplied equation, not additional verified NEET PYQs.

What are the slope and intercept of the natural-log plot?

Related practice question 1: For this reaction, the slope is negative and the intercept is 14.34. For a plot of the natural logarithm of the rate constant against reciprocal temperature, compare with the straight-line form:

y = m x + c, y = \ln k, x = \frac{1}{T}

The supplied equation gives:

m = - 1.25 \times 10^{4} K, c = 14.34

Recover the activation energy from the slope:

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

E_{a} = - (- 1.25 \times 10^{4} K) (1.987 c a l m o l^{- 1} K^{- 1}) \approx 24.84 k c a l m o l^{- 1}

The two negative signs cancel. The activation energy is positive.

How does using common logarithms change the equation and activation energy?

Related practice question 2: Divide the entire supplied equation by 2.303 to express it using common logarithms. The activation energy stays unchanged because the common-logarithm Arrhenius form contains the same conversion factor. $\ln k = 2.303 \log_{10} k$

\log_{10} k = \frac{14.34}{2.303} - (\frac{1.25 \times 10^{4}}{2.303}) (\frac{1}{T})

Compare this with:

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

Matching the coefficient gives:

E_{a} = 2.303 R (\frac{1.25 \times 10^{4} K}{2.303}) = (1.25 \times 10^{4} K) R

E_{a} \approx 24.84 k c a l m o l^{- 1}

Before selecting an answer, identify the logarithm base, match the coefficient, and convert the energy units.

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Frequently asked questions

What is the answer to the Arrhenius Equation NEET 2026 question?

For ln k = 14.34 − (1.25 × 10^4)/T, the activation energy is 24.84 kcal/mol, so option A is correct. Comparing with ln k = ln A − Ea/(RT) gives Ea/R = 12,500 K. Multiplying by R = 1.987 cal mol⁻¹ K⁻¹ gives 24,837.5 cal/mol, which is 24.8375 kcal/mol.

Why is 14.34 not the activation energy?

The constant 14.34 represents ln A, the natural logarithm of the Arrhenius prefactor. The coefficient of 1/T represents Ea/R, so activation energy comes from multiplying 12,500 K by the gas constant, not from the intercept.

When do I use 2.303 in the Arrhenius equation?

Use the factor 2.303 when working with common logarithms: log₁₀ k = log₁₀ A − Ea/(2.303RT). No factor of 2.303 is needed for the supplied natural-logarithm equation. Converting the entire equation from natural to common logarithms leaves the activation energy unchanged.

How do I find activation energy from an Arrhenius plot?

For a plot of ln k against 1/T, the slope m equals −Ea/R, so Ea = −mR. The supplied equation has slope −12,500 K and intercept 14.34. Using the supplied gas constant gives a positive activation energy of approximately 24.84 kcal/mol.

Arrhenius Equation NEET 2026: Activation Energy Solution

What is the answer to the Arrhenius Equation NEET 2026 question?

How do you match the given equation to the Arrhenius form?

How do you calculate activation energy and convert cal to kcal?

Why does choosing 14.34 give the wrong answer?

How do you solve related graph and common-logarithm questions?

What are the slope and intercept of the natural-log plot?

How does using common logarithms change the equation and activation energy?

Frequently asked questions

What is the answer to the Arrhenius Equation NEET 2026 question?

Why is 14.34 not the activation energy?

When do I use 2.303 in the Arrhenius equation?

How do I find activation energy from an Arrhenius plot?

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