12) let the rate constant of both rxns at 25 deg. be K
let temp. coeff of rxn. 1 = r then coeff of 2nd rxn = 2.5 r
then by def. of temp. coeff. rate constants will become r^7 * k and (2.5r)^7 * k
hence ans. = (2.5)^7
10 Answers
1
213)(A)
In reactions no. of equivalents remain the same.
3/(x+8)=5/(x+35.5)
x=33.25(ans)
13) MOn/2 ---> M Cln
=> 3M + 8n = 5M+(35.5)n
=> solving we get M/n = 33.25 (A)
212)Let temp. coefficient be x.
Temp. diff=θ=71
r1final=r1(1+x*θ)
r2final=r2(1+2.5x*θ)
r2/r1=(1+2.5x*71)/(1+x*71)≈2.5(ans)
21Thanks aditya for correction.
Resistance increases linearly with temp, ie why the formula i used is valid only incase of linear dependance.
However for kinetics the rate is exponentially dependent on temp.
71So what is the answer for second?
THanks all for the rest.
1has to be slightly greater than (2.5)^7 as i already posted.
since i've approximated 96 degrees to 95 degrees.