The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy =Ea27.0/kJmol. If the rate constant of this reaction is ×1.8103·M−1s−1 at 24.0°C, what will the rate constant be at −35.0°C?Round your answer to 2 significant digits.
Question
The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy =Ea27.0/kJmol. If the rate constant of this reaction is ×1.8103·M−1s−1 at 24.0°C, what will the rate constant be at −35.0°C?Round your answer to 2 significant digits.
Solution
The Arrhenius equation is given by:
k = A * exp(-Ea / RT)
where: k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin.
Given: Ea = 27.0 kJ/mol = 27000 J/mol k1 = 1.81 x 10^3 M^-1 s^-1 at T1 = 24.0°C = 297.15 K We need to find k2 at T2 = -35.0°C = 238.15 K
We can use the two-point form of the Arrhenius equation, which is:
ln(k2/k1) = -Ea/R * (1/T2 - 1/T1)
Rearranging for k2 gives:
k2 = k1 * exp[-Ea/R * (1/T2 - 1/T1)]
Substituting the given values and using R = 8.314 J/mol*K, we get:
k2 = 1.81 x 10^3 * exp[-27000/(8.314) * (1/238.15 - 1/297.15)]
Calculating the above expression will give the value of k2. Remember to round your answer to 2 significant digits.
Similar Questions
The rate constant k for a certain reaction is measured at two different temperatures:temperature k392.0°C ×1.71011470.0°C ×2.91011Assuming the rate constant obeys the Arrhenius equation, calculate the activation energy Ea for this reaction.Round your answer to 2 significant digits.
The rate constant, the activation energy and the Arrhenius parameter of a chemical reaction at 25∘C are 3.0 × 10−4s−1, 104.4KJmol&6 × 1014s−1 respectively. The value of rate constant as T − ∞ ,
A certain catalyzed reaction is known to have an activation energy =Ea17.0/kJmol. Furthermore, the rate of this reaction is measured at 362.K and found to be ×3.5/105Ms. Use this information to answer the questions in the table below.Suppose the concentrations of all reactants is kept the same, but the temperature is raised by 5% from 362.K to 380.K. The rate will How will the rate of the reaction change?Suppose the concentrations of all reactants is kept the same, but the catalyst is removed, which has the effect of raising the activation energy by 10%, from 17.0/kJmol to 18.7/kJmol. The rate will How will the rate of the reaction change?
A certain catalyzed reaction is known to have an activation energy =Ea50.0/kJmol. Furthermore, the rate of this reaction is measured at 310.K and found to be ×3.8/101Ms. Use this information to answer the questions in the table below.
Some measurements of the initial rate of a certain reaction are given in the table below.N2 H2 initial rate of reaction0.935M 1.66M /40.0Ms0.935M 0.484M /3.40Ms0.120M 1.66M /0.659MsUse this information to write a rate law for this reaction, and calculate the value of the rate constant k.Round your value for the rate constant to 2 significant digits. Also be sure your answer has the correct unit symbol.
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.