A certain reaction has an activation energy of 66.58 kJ/mol. At what Kelvin temperature will the reaction proceed 4.50 times faster than it did at 289 K
given:
Ea= 66.58 kj/mol=66,580 j/mol
T1=298.15k
K2=4.5 (K1)
find: T2
solution:
ln (k2/k1)= Ea/R (1/T1-1/T2)
ln (4.5 k1/k1) =Ea/R ( 1/298.15k-1/T2)
T2=316 K
Hope that helps 👍
To find the Kelvin temperature at which the reaction will proceed 4.50 times faster than it did at 289 K, we can use the Arrhenius equation:
k2 = k1 * exp((Ea / R) * (1/T2 - 1/T1))
Where:
k2 = rate constant at the desired temperature (unknown)
k1 = rate constant at 289 K (known)
Ea = activation energy in Joules (known)
R = gas constant = 8.314 J/(mol·K)
T2 = desired temperature in Kelvin (unknown)
T1 = 289 K (known)
First, let's convert the activation energy from kJ/mol to Joules/mol:
Ea = 66.58 kJ/mol * 1000 J/kJ
Ea = 66,580 J/mol
Now we can rearrange the equation to solve for T2:
k2/k1 = exp((Ea / R) * (1/T2 - 1/T1))
Taking the natural logarithm of both sides:
ln(k2/k1) = (Ea / R) * (1/T2 - 1/T1)
We know that k2 = 4.50 * k1, so we can substitute it in:
ln(4.50 * k1 / k1) = (Ea / R) * (1/T2 - 1/T1)
ln(4.50) = (Ea / R) * (1/T2 - 1/T1)
Now we can substitute the values we know:
ln(4.50) = (66,580 J/mol / (8.314 J/(mol·K))) * (1/T2 - 1/289 K)
Simplifying:
ln(4.50) = 8,007.8 (1/T2 - 1/289 K)
Now we can isolate the temperature term:
1/T2 - 1/289 K = ln(4.50) / 8,007.8
1/T2 = ln(4.50) / 8,007.8 + 1/289 K
Now solve for T2:
T2 = 1 / (ln(4.50) / 8,007.8 + 1/289 K)
Calculating T2 gives us the temperature at which the reaction will proceed 4.50 times faster than at 289 K.