Suppose a substance A decompose into P and Q is parallel paths with rate constan

Question

Suppose a substance A decompose into P and Q is parallel paths with rate constants given by k_p = (10^15 s^-1)e^-126,000 k_0 = (10^14 s^-1)e^where the activation energies are given in J mol^-1 (R = 8.314 J K^-1) is each path a first-order, second order or third-order reaction? At what temperature will both products be formed at the same rate? i)second-order, ii) 100 K i) third-order first-order, ii) 1105 K i)first-order ii)1105 K i)first-order ii)2540 K i)second-order, ii) 1105 Consider the first-order gas phase decomposition of a peroxide with rate constants k_100 degree C = 4.3 times 10^-7 s^-1 and k_200 degree C = 2.9 times 10^-2 s^-1. Choose the closest answer. What is Delta G^0t for the activation reaction at 100 degree C? 250 kJ/mol 138 kJ/mol 53 kJ/mol 317 kJ/mol 0.031 kJ/mol

Explanation / Answer

We know the equation that relates deltaG0 and equilibrium constant K is :

deltaG0 = -2.303 RT logK

By substituting the given values in the above eqaution, we get:

deltaG0 = - 2.303 x 8.314J.k-1 mol-1 x (100+273)K x log (4.3x10^-7)

             = 54517. 3 J/mol

             = 54.517kJ/mol

nearest option is 53kJ/mol, option C.

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