GATE 2023 BT Q.47: solution and answer

GATE 2023 · BTQ.47 · 2 marks · NAT

Temperature of a reaction with an activation energy value of $15 kcal mol^{- 1}$ is increased from $300 K$ to $310 K$ . If the value of the ideal gas constant ( $R$ ) is $1.9872 cal mol^{- 1} K^{- 1}$ , the ratio of the reaction rate constants $(k_{310} / k_{300})$ (rounded off to two decimal places) is ________.

See the question in the paper

Tap to open full size

Answer and solution

ANSWER

$k_{310} / k_{300} = e^{0.8116} \approx 2.25$ . A 10 K rise roughly doubles the rate constant, and a bit more.

Official answer: 2.24 to 2.26Source: Official answer key, GATE 2023 BT (IIT Kanpur)

Arrhenius equation: $k = A e^{- E_{a} / R T}$ , where $k$ is the rate constant, $A$ the pre-exponential factor, $E_{a}$ the activation energy, $R$ the gas constant and $T$ the absolute temperature. Write it at $T_{1} = 300 K$ and $T_{2} = 310 K$ and divide one by the other. $A$ cancels (it is taken as independent of temperature): $ln \frac{k _{310}}{k _{300}} = \frac{E _{a}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})$
Put $E_{a}$ and $R$ in the same energy unit: $E_{a} = 15 kcal mol^{- 1} = 15000 cal mol^{- 1}$ , so $\frac{E _{a}}{R} = \frac{15000 cal mol ^{- 1}}{1.9872 cal mol ^{- 1} K ^{- 1}} = 7548.31 K$ .
Temperature term: $\frac{1}{300} - \frac{1}{310} = \frac{310 - 300}{300 \times 310} = \frac{10}{93000} = 1.0753 \times 1 0^{- 4} K^{- 1}$ .
Multiply: $ln \frac{k _{310}}{k _{300}} = 7548.31 K \times 1.0753 \times 1 0^{- 4} K^{- 1} = 0.8116$ . The result has no units, as it must.
Take the exponential: $\frac{k _{310}}{k _{300}} = e^{0.8116} = 2.2516 \approx 2.25$ .

KEY IDEA

Divide the Arrhenius equation at one temperature by the same equation at the other. The pre-exponential factor cancels, leaving $ln (k_{2} / k_{1}) = \frac{E _{a}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})$ . Only $E_{a}$ , $R$ and the two absolute temperatures are needed, provided $E_{a}$ and $R$ use the same energy unit.

WATCH OUT FOR

Mixing kcal and cal. Putting $E_{a} = 15$ over $R = 1.9872$ gives a ratio of about 1.00. Also watch the sign and order of $(\frac{1}{T _{1}} - \frac{1}{T _{2}})$ : reversing it gives the inverse ratio, about 0.44. Approximating the bracket as $Δ T / T^{2}$ with $T = 300 K$ inflates the answer to about 2.31. Finally, rounding $R$ to 2 gives 2.2400, which only just sits on the edge of the key, so keep $R = 1.9872$ as printed.

From the GATE 2023 BT paper. Every question of this paper · Download the full paper · All 9 Enzyme kinetics questions, with notes

GATE 2023 BT Q.47: solution and answer

Answer and solution

More Enzyme kinetics questions