Question

Answer
It is given that ( T_{1}=298 mathrm{K} ) ( therefore T 2=(298+10) mathrm{K} )
( =308 mathrm{K} )
We also know that the rate of the reaction doubles when temperature is increased by ( 10^{circ} ). Therefore, let us take the value of ( mathrm{k}_{1}=mathrm{k} ) and that of ( mathrm{k}_{2} )
( =2 mathrm{k} )
( mathrm{Also}, mathrm{R}=8.314 mathrm{J} mathrm{K}^{-1} mathrm{mol}^{-1} )
Now, substituting these values in the equation:
( log frac{k_{2}}{k_{1}}=frac{E_{2}}{2.303 R}left[frac{T_{2}-T_{1}}{T_{1} T_{2}}right] )
We get:
( log frac{2 k}{k}=frac{E_{s}}{2.303 times 8.314}left[frac{10}{298 times 308}right] )
( Rightarrow log 2=frac{E_{0}}{2.303 times 8.314}left[frac{10}{298 times 308}right] )
( Rightarrow E_{a}=frac{2.303 times 8.314 times 298 times 308 times log 2}{10} )
( =52897.78 mathrm{Jmol}^{-1} )
( =52.9 mathrm{kJ} mathrm{mol}^{-1} )

# Question 8. The rate constant of the chemical reaction doubled for an increase of 10 K in absolute temperature from 295 K. Calculate E. Auchane nuation locka E, [11] Archani anation

Solution