Question

A certain reaction has an activation energy of 54.0 kJ/mol. As the temperature is increased from 22C to a higher temperature, the rate constant increases by a factor of 7.00. Calculate the higher temperature.

Answer 1

The higher temperature will be:

As the temperature is increased from 22°C to a higher temperature, the rate constant increases by a factor of 7.00. The final temperature will be 271.74K.

What is the activation energy?

Activation energy is defined as the extra energy provided required to activate the molecules to start reacting and forming intermediate and later product.

Arrhenius equations given the relation rate constant and activation energy.

$LnK=Ae^{-\frac{E_a}{RT}$

Where,

K= Rate constant

A= Pre exponential factor

$E_a=$ Activation energy

When two rate constants are given $K_1\,$ and $\,K_2$

$ln\frac{K_2}{K_1}=\frac{E_a}{R}(\frac{1}{T_2}-\frac{1}{T_1})$

Given,

$T_1$ = 22°C = 295K

Rate constant$(K_1)$ increases by a factor of 7,i.e, $K_2=7K_1$

$E_a$= 54 kJ/mole = 54000 J/mole

R= 8.314 $J\,mol^{-1}K^{-1}$

Putting the values,

$ln\frac{K_2}{K_1}=\frac{E_a}{R}(\frac{1}{T_2}-\frac{1}{T_1})$

ln 7 x 8.31454000 = (1/$T_{2}$ - 1/295)

0.299 x $10^{-3}$ + 1/295 = 1/$T_{2}$

0.00368 = 1/$T_{2}$

$T_{2}$ = 1/0.00368

$T_{2}$ = 271.74 K

Therefore,

The increase in temperature will change from 295K to 271.74K.

Learn more about Arrhenius equation of activation here,

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