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Q.(i) Define the order of reaction.

(1)
(ii) The rate constants of a reaction at 500 K and 700 K are 0.025 sec-1 and 0.075 sec-1 respectively. Calculate the value of Ea and A. (2)
Himachal HpboseHPBOSE Plus Two Board 2026Subjective· 3mImportance★★★★★
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(i) Order of reaction is the sum of exponents of concentration in the empirical rate law. (ii) Using the two-temperature Arrhenius equation with k₁ = 0.025 s⁻¹ at 500 K and k₂ = 0.075 s⁻¹ at 700 K gives Ea ≈ 16.0 kJ/mol and pre-exponential factor A ≈ 1.17 s⁻¹.

(i) Order of reaction

The order of a reaction is defined as the sum of the powers (exponents) to which the concentration terms of the reactants are raised in the experimentally determined rate law. For a reaction with rate =k[A]x[B]y= k[A]^x[B]^y, the overall order is x+yx + y. Unlike molecularity, order is found experimentally and can be zero, a fraction, or a whole number.

(ii) Finding Ea and A

Given: k1=0.025 s−1k_1 = 0.025\ s^{-1} at T1=500 KT_1 = 500\ K; k2=0.075 s−1k_2 = 0.075\ s^{-1} at T2=700 KT_2 = 700\ K.

Use the two-point Arrhenius (Van't Hoff-type) equation:

ln⁡k2k1=EaR(1T1−1T2)\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)

Step 1: k2k1=0.0750.025=3⇒ln⁡3=1.0986\dfrac{k_2}{k_1} = \dfrac{0.075}{0.025} = 3 \Rightarrow \ln 3 = 1.0986

Step 2: 1T1−1T2=1500−1700=0.002−0.0014286=5.714×10−4 K−1\dfrac{1}{T_1} - \dfrac{1}{T_2} = \dfrac{1}{500} - \dfrac{1}{700} = 0.002 - 0.0014286 = 5.714\times10^{-4}\ K^{-1}

Step 3: Solve for Ea (R = 8.314 J K⁻¹ mol⁻¹):

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