General Chem problem

2) 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

3)(A)
In reactions no. of equivalents remain the same.
3/(x+8)=5/(x+35.5)
x=33.25(ans)

3) MO_n/2 ---> M Cl_n

=> 3M + 8n = 5M+(35.5)n
=> solving we get M/n = 33.25 (A)

lol didn't see swaraj's post.

2)Let temp. coefficient be x.
Temp. diff=θ=71
r1_final=r1(1+x*θ)
r2_final=r2(1+2.5x*θ)
r2/r1=(1+2.5x*71)/(1+x*71)≈2.5(ans)

@swaraj - temp. coeff = K_T+10^oK_T (K → Rate constant)

has to be slightly greater than (2.5)^7 as i already posted.
since i've approximated 96 degrees to 95 degrees.