The equilibrium constant for a certain reaction decreases from 8.84 to 3.25*10^-2 when the temperature increases from 25 degrees celsius to 75 degree celsius. Estimate the temperature where K= 1.00...

The equilibrium constant for a certain reaction decreases from 8.84 to 3.25*10^-2 when the temperature

increases from 25 degrees celsius to 75 degree celsius. Estimate the temperature where K= 1.00 for this reaction. 

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jerichorayel | College Teacher | (Level 2) Senior Educator

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In case of variation of equilibrium constant with temperature, we use the van't Hoff equation which states that:

`ln((k_2)/(k_1)) = (-DeltaH)/R (1/(T_2) - 1/(T_1))`

The first step is to get the value of Delta H. 

k2 = 3.25x10^-2    T2 = 75+ 273.15 = 348.15 K

k1 = 8.84               T1 = 25 + 273.15 = 298.15 K

R = gas constant = 8.314 J/mol-K

`ln((3.25x10^-2)/(8.84)) = (-DeltaH)/(8.314) (1/(348.15) - 1/(298.15))`

`-5.6058 =(-DeltaH)/(8.314) (-4.8169x10^-4)`

`-Delta H = ((-5.6058)(8.314))/(-4.8169x10^-4)`

`(-Delta H = 96, 756.24 J/mol) -1`

`Delta H` = -96, 756.24 J/mol

When K = 1.00, we just change the parameter of either the K1 or K2. In this case we change the value of K2.

`ln((k_2)/(k_1)) = (-DeltaH)/R (1/(T_2) - 1/(T_1))`

k2 = 1.00               T2 = ?

k1 = 8.84               T1 = 25 + 273.15 = 298.15 K

R = gas constant = 8.314 J/mol-K

`DeltaH`  = -96, 756.24 J/mol

`ln((1.00)/(8.84)) = (-(-96, 756.24))/(8.314) (1/(T_2) - 1/(298.15))`

`-2.1793 = 11637.75 (1/(T_2) - 3.354x10^-3)`

`1/(T_2) = (-2.1793)/(11637.75) +3.354x10^-3`

`1/(T_2) = 3.54126x10^-3`

`T_2` = 315.78 - 273.15 = 42.63 degrees Celsius


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