Question

the slope of a line graph of ln k versus 1/T is -1.82 ×10^4 K what is the final temperature if it is observed that the rate constant of a reaction increases by a factor of 20 when increased from 25 deg Celsius to this higher final temperature ?
(detailed solution please )

Answer 1

is a lineal comportament so Y = mX + b where

Y = K, X = 1/T and m = -1.82x10⁴K so

X = 1/(T + 25C) as the lineal is Y =  -1.82x10⁴K/T and Y2 =   -1.82x10⁴K x 1/(T +25C) as   Y2 = 20Y

20Y =   -1.82x10⁴K x 1/(T + 25C)

now have a sisten whit 2 incongnite and two ecuations using a elimination methode

1 )   Y =  -1.82x10⁴K/T =)  -20( Y =  -1.82x10⁴K/T ) =) -20Y = 3.64x10⁵K/T

2) 20Y =   -1.82x10⁴K x 1/(T + 25C)

now its a simple math problems so as T is in demonator invert all equation and rewrite so

1/-20Y = 1/3.64x10⁵K/T = -1/20Y = T/3.64x10⁵K

1/20Y = 1/-1.82x10⁴K x 1/(T + 298K) = 1/20Y = -(T + 25C) /1.82x10⁴K

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0 = T(1/3.64x10⁵K  - 1/1.82x10⁴K) - 25C/1.82x10⁴K solving

1.37x10^-3C = T ( -5.22x10^-5) =) T = 1.37x10^-3C/ -5.22x10^-5 = -26.24C

the initial T is -26.24C and the final Tf = -26.24C + 25C = -1.24C

Tf = -1.24C