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Q.For a reaction, the first-order rate constant at 600 K is 1.6×10−5 s−11.6 \times 10^{-5}\ s^{-1}. If the activation energy is 209 kJ mol−1209\ \text{kJ mol}^{-1}, calculate the rate constant of the reaction at 700 K.

Meghalaya MboseMBOSE Meghalaya Intermediate Board 2020Subjective· 2mImportance★★★★★
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The Arrhenius two-point form lets us find kk at a new temperature once EaE_a and kk at one temperature are known. Here raising T from 600 K to 700 K increases the rate constant by a factor of about 400.

Arrhenius equation (two-temperature form):

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)

Data: k1=1.6×10−5 s−1k_1 = 1.6\times10^{-5}\ s^{-1} at T1=600 KT_1 = 600\,K; Ea=209 kJ mol−1=2.09×105 J mol−1E_a = 209\ kJ\,mol^{-1} = 2.09\times10^{5}\,J\,mol^{-1}; R=8.314 J mol−1K−1R = 8.314\,J\,mol^{-1}K^{-1}; find k2k_2 at T2=700 KT_2 = 700\,K.

Step 1 — the temperature term:

1T1−1T2=1600−1700=700−600600×700=100420000=2.381×10−4 K−1\frac{1}{T_1}-\frac{1}{T_2} = \frac{1}{600}-\frac{1}{700} = \frac{700-600}{600\times700} = \frac{100}{420000} = 2.381\times10^{-4}\,K^{-1}

Step 2 — plug in:

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