Q.74. For reaction A → P, rate constant k = 1.5 s⁻¹ at 27°C. If activation energy for the above rea

Q: What is the correct answer for this exam Chemistry Q74?

The correct answer is 41. Sol. We can use the Arrhenius equation in its integrated form for two different temperatures: $\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)$ Given values: $k_1 = 1.5 \times 10^{-3}$ s⁻¹ at $T_1 = 27°C = 27 + 273 = 300$ K $k_2 = 4.5 \times 10^{-3}$ s⁻¹ at $T_2$ (unknown) $E_a = 60$ kJ mol⁻¹ = $60 \times 10^3$ J mol⁻¹ $R = 8.3$ J K⁻¹ mol⁻¹ Substitute t

Q: Q.74. For reaction A → P, rate constant k = 1.5 s⁻¹ at 27°C. If activation energy for the above reaction is 60 kJ mol⁻¹, then the temperature (in °C) at which rate constant, k = 4.5 s⁻¹ is…

The correct answer is 41. Sol. We can use the Arrhenius equation in its integrated form for two different temperatures: $\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)$ Given values: $k_1 = 1.5 \times 10^{-3}$ s⁻¹ at $T_1 = 27°C = 27 + 273 = 300$ K $k_2 = 4.5 \times 10^{-3}$ s⁻¹ at $T_2$ (unknown) $E_a = 60$ kJ mol⁻¹ = $60 \times 10^3$ J mol⁻¹ $R = 8.3$ J K⁻¹ mol⁻¹ Substitute t

Question

Q.74. For reaction A → P, rate constant k = 1.5  s⁻¹ at 27°C. If activation energy for the above reaction is 60 kJ mol⁻¹, then the temperature (in °C) at which rate constant, k = 4.5  s⁻¹ is _________ (Nearest integer)
Given: log 2 = 0.30, log 3 = 0.48, R = 8.3 J K⁻¹ mol⁻¹, ln 10 = 2.3

FirstInTest · Accepted Answer

Sol. We can use the Arrhenius equation in its integrated form for two different temperatures:
$\ln\left(\frac{k_2}{k_1}ight) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}ight)$

Given values:
$k_1 = 1.5 	imes 10^{-3}$ s⁻¹ at $T_1 = 27°C = 27 + 273 = 300$ K
$k_2 = 4.5 	imes 10^{-3}$ s⁻¹ at $T_2$ (unknown)
$E_a = 60$ kJ mol⁻¹ = $60 	imes 10^3$ J mol⁻¹
$R = 8.3$ J K⁻¹ mol⁻¹

Substitute the values into the equation:
$\ln\left(\frac{4.5 	imes 10^{-3}}{1.5 	imes 10^{-3}}ight) = \frac{60 	imes 10^3 	ext{ J/mol}}{8.3 	ext{ J K⁻¹ mol⁻¹}} \left(\frac{1}{300 	ext{ K}} - \frac{1}{T_2}ight)$

$\ln(3) = \frac{60000}{8.3} \left(\frac{1}{300} - \frac{1}{T_2}ight)$

Using $\ln 3 = 2.3 	imes \log 3 = 2.3 	imes 0.48 = 1.104$

$1.104 = \frac{60000}{8.3} \left(\frac{1}{300} - \frac{1}{T_2}ight)$

$1.104 = 7228.915 \left(\frac{1}{300} - \frac{1}{T_2}ight)$

$\frac{1.104}{7228.915} = \frac{1}{300} - \frac{1}{T_2}$

$0.0001527 = 0.003333 - \frac{1}{T_2}$

$\frac{1}{T_2} = 0.003333 - 0.0001527$

$\frac{1}{T_2} = 0.0031803$

$T_2 = \frac{1}{0.0031803} \approx 314.43$ K

Convert $T_2$ to °C:
$T_2(°C) = T_2(K) - 273 = 314.43 - 273 = 41.43°C$

Rounding to the nearest integer, the temperature is 41°C.

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