Question

If a temperature increase from 19.0 C to 33.0 C triples the rate constant for a reaction, what is the value of the activation barrier for the reaction?

Answer 1

Ok this is easy. From arrehinus equation get the relation for rate constants and temperature.

And just substitute values

Please mark as Brainliest if you think it helped you,

Thanks,

EndlessGamer

Answer 2

Answer:

The activation energy barrier is 57.1 kJ/mol.

Explanation:

The Arrhenius equation gives the relation between temperature and reaction rates:

#color(blue)(|bar(ul(color(white)(a/a) k = Ae^(-E_"a"/(RT))color(white)(a/a)|)))" "#

where

#k# = the rate constant

#A# = the pre-exponential factor

#E_"a"# = the activation energy

#R# = the Universal Gas Constant

#T# = the temperature

If we take the logarithms of both sides, we get

#lnk = lnA - E_"a"/(RT)#

Finally, if we have the rates at two different temperatures, we can derive the expression

#color(blue)(|bar(ul(color(white)(a/a) ln(k_2/k_1) = E_"a"/R(1/T_1 -1/T_2)color(white)(a/a)|)))" "#

In your problem,

#T_2 =(33.0 + 273.15) K = 306.15 K#

#T_1 = (19.0 + 273.15) K = 292.15 K#

#k_2/k_1 = 3#

Now, let's insert the numbers.

#ln(k_2/k_1) = E_"a"/R(1/T_1 -1/T_2)#

#ln3 = E_"a"/("8.314 J"·color(red)(cancel(color(black)("K")))^"-1""mol"^"-1") (1/(292.15 color(red)(cancel(color(black)("K")))) - 1/(306.15 color(red)(cancel(color(black)("K")))))#

#ln3= E_"a"/("8.314 J·mol"^"-1") × 1.565 × 10^"-4"#

#E_a = (ln3 × "8.314 J·mol"^"-1")/(1.565 × 10^"-4") = "57 100 J/mol" = "57.1 kJ/mol"#