Question

The rate constant $K_{1}$ of a reaction is found to be double that of rate constant $K_{2}$ of another reaction. The relationship between corresponding activation energies of the two reactions at same temperature ( $E_{1}$ and $E_{2}$ ) can be represented as

• A. $E_{1}>E_{2}$
• B. $E_{1} < E_{2}$
• C. $E_{1}=E_{2}$
• D. None of these

Answer 1

Answer: (D)

$K_{1}=A_{1}e^{- E_{1} / R T}$

and $K_{2}=A_{2}e^{- E_{2} / R T}$

$\frac{K_{1}}{K_{2}}=\frac{A_{1}}{A_{2}}e^{\left(\right. - E_{1} + E_{2} \left.\right) / R T}$

As $A_{1}$ and $A_{2}$ are not given here so none of the relation is correct.