4 a) (6 points) In the 1960’s the following mechanism for reaction of H2(g) with

Question

4 a) (6 points) In the 1960’s the following mechanism for reaction of H2(g) with I2(g) to form HI(g) was proposed.

I2(g) 2I(g) fast RXN 1

H2(g) + I(g) H2I(g) fast RXN 2

H2I(g) +I(g) 2HI(g) slow RXN 3

The experimental rate law was determined to be rate = k [H2][I2]. Is this mechanism correct?

b) (4 points) Using the above mechanism, and the activation energies of 1, 0.5 and 75 kJ mol-1, for steps 1, 2 and 3 respectively, determine the temperature required to increase the rate of the reaction by a factor of 10 relative to the rate at 298 K.

I just need help with part b. I am not sure how to get the activation energy to use.

Explanation / Answer

Activation energy is 75 KJ/mol ( we consider activation energy of slowest step)

now we have formula

ln ( k2/k1) = ( Ea/R) ( 1/T1 -1/T2)

where T1 = 298 , , we need to find T2 susch that k2/k1 = 10 i.e rate increases 10 times

Ea = 75 KJ/mol = 75000 J/mol , R= gas constant = 8.314 J/molK

now ln ( 10) = ( 75000/8.314) ( 1/298 -1/T2)

T2 = 322.5 K

thus at temp 322.5 K our rate is doubled i.e k2 = 10 times k1.

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