Question

A certain reaction is known to have an activation energy E = 70.0 kJ/mol. Furthermore, the rate of this reaction is measured at 362. K and found to be 7.9x 10 MIs. Use this information to answer the questions in the table below. (You may need to use the scrollbar to see all the choices.) Suppose the concentrations of all reactants is kept the same, but the temperature is lowered by 10% from 362' K to 326. K. The rate will How will the rate of the reaction change? Suppose the concentrations of all reactants is kept the same, but the catalyst is added, which has the effect of lowering the activation energy by 5%, from 70.0 kJ/mol to 66.5 kJ/mol. The rate will choose one How will the rate of the reaction change?

Answer 1

Activation energy E_a = 70 K J/m o l = 70, 00 J/mole Initial Temperature T_1 = 362 k Initial Rate constant k_1 = 7.9 times 10^-3 m/sec Final Temperature T_2 = 326 k let final rate constant k_2 we know, l n k_2/k_1 = E_a/R (T_2 -T_1)/T_1 T_2 or, l n k_2/k_1 = 70000/8.314 times (326 -362)/362 times 326 or, l n k_2/k_1 = -2.568 or, k_2 = k_1 times e^-2.568 = 7.9 times 10^-3 times e^-2.568 or k_2 = 6.055 times 10^-4 k_2 < k_1 Therefore Rate constant value decreases The rate will be slow than the previous rate \r\n E_a = 66.5 K J/m o l = 66500 J/mole k_1 = 7.9 times 10^-3 T_1 = 362 k As the activation energy