Question

Consider the graph for the reaction:

\[\ce{H2 + I2 {⇌} 2HI}\]

1. Calculate enthalpy of reaction and activation energy for the backward reaction.
2. How will the catalyst affect the rate of this reaction? Explain.

Answer 1

I. Enthalpy of reaction:

`(Δ"H") = "E"_("products") - "E"_"reactants"`

= 100 − 90

= +10 kJ mol⁻¹

Activation energy for backward reaction:

`"E"_"a"​("back") = "E"_"activated complex"​ - "E"_"products"`

​= 140 − 100

= 40 kJ mol⁻¹

II. Catalyst will increase the rate of reaction, as the activation energy required to form an intermediate activated complex between the reactant and catalyst is lower than the activation energy required for forming a complex without a catalyst.