Question

Consider the following reaction which follows zero-order reaction kinetics

          A(g) + Catalyst(s) ? B(g)                ?H = -226.9 kJ

Notice that the reaction is exothermic, so in a closed system it is possible for this to get out of hand: The reaction releases energy, which increases the temperature, which increases the rate, which makes it heat up faster, which makes it react faster

Answer 1

For a zero order reaction , k = [A]_o-[A]_t/t

Given that the heat sink absorbs heat at 160 kJ/min or 2.67 kJ/s,

delta H for the raction is 226.9 kJ/mol. To obtain 2.67 kJ the no. of mole of A to be consumed per sec should be, 2.67/226.9 = 0.01175 moles.

Thus if k doesnt exceed 0.01175 or 11.75 mM/s, then the system will not be a thermal hazard.

From the table, we can find Ea/R for the raction using

ln(k1/k2) = (Ea/R){(1/T1)-(1/T2)} Use the values of ks at T = 35 and 45 C, we get Ea/R = -3346.16

Again, using the same equation, we can calculate the temperature at which the k = 11.75 mM/s, which is

1/328 - 1/T = R/Ea*ln(4.109/11.75) = 3.14x10^-4

T = 365.66 K or 92.66 ^oC

Therefore, the temperature should not exceed 92 ^oC