General Chem problem

Solution Please!

10 Answers

71
Vivek @ Born this Way ·

1
rishabh ·

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

21
Swaraj Dalmia ·

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

1
rishabh ·

3) MOn/2 ---> M Cln

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

1
rishabh ·

lol didn't see swaraj's post.

21
Swaraj Dalmia ·

2)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)

262
Aditya Bhutra ·

@swaraj - temp. coeff = KT+10oKT (K → Rate constant)

21
Swaraj Dalmia ·

Thanks 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.

71
Vivek @ Born this Way ·

So what is the answer for second?

THanks all for the rest.

1
rishabh ·

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

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