What is the value of delta H for a reaction the value of K doubles when the temerature is increased from 25 deg C to 35 deg C?
The problem provides the condition that `k_35 = 2k_25` , hence you should come up with the substitution `k_35 = k_1` and `k_25 = k_2` .
You may use the following equation such that:
`ln k_1 = ln A - E_a/(RT_1)`
`ln k_2 = ln A - E_a/(RT_2)`
You need to subtract `ln k_2 = ln A - E_a/(RT_2)` from `ln k_1 = ln A - E_a/(RT_1)` such that:
`ln k_1 - ln k_2= ln A - E_a/(RT_1) - ln A+ E_a/(RT_2)`
Using the laws of logarithms and reducing like terms yields:
`ln(k_1/k_2) = (E_a/R)(1/T_2 - 1/T_1)`
`E_a = (RT_1*T_2)/(T_2-T_1)*ln(k_1/k_2)`
You need to convert Celsius degrees to Kelvin such that:
`T_1 = 25 + 273 = 298`
`T_2 = 35 + 273 = 308`
`E_a = (8.314*298*308)/(308-298)*ln(2k_2/k_2)`
`E_a = 76309.2176*0.6931`
`E_a = 52889.918`
`E_a ~~ 53 (kJ)/(mol)`
Hence, evaluating the energy of activation yields entalpy `Delta H = 53 (kJ)/(mol)` .