Question

Consider the reaction 2HI(g)→H2(g)+I2(g). At 585 K, the rate constant is 9.64×10−5Lmol s. At 690. K,...

Consider the reaction 2HI(g)→H2(g)+I2(g). At 585 K, the rate constant is 9.64×10−5Lmol s. At 690. K, the rate constant is 2.83×10−3Lmol s. Use the Arrhenius equation to calculate the activation energy for the reaction. Ea=−R[lnk2−lnk1(1T2)−(1T1)] Provide your answer below:

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Answer #1

first find the values at different temperatures and substitute the values in the equation
Ea=−R[lnk2−lnk1(1T2)−(1T1)] or the arrhenius equation can also be represented as follows

=> log K2 / K1 = Ea / 2.303*R(1/T1 - 1/T2)

Where k1 = 9.64*10-5   and T1 = 585 K

k2 = 2.83*10-3 and T2 = 690 k

R = 8.314

substituting the values in the above equation

log(2.83*10-3/ 9.64*10-5 ) = Ea / 2.303 * 8.314 (1/585 -1/690)

solving for Ea => 108033.80 j/mol

                         => 108.033 kj/ mol

answer => activation energy Ea = 108.033 kj/ mol

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Consider the reaction 2HI(g)→H2(g)+I2(g). At 585 K, the rate constant is 9.64×10−5Lmol s. At 690. K,...
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