Answer:
1(a) Second order
1(b) 44.0 minutes
2(a) Ea, forward = 85.6 kJ/mol
Ea, reverse = 60.1 kJ/mol
3) k = 9.47 M⁻¹s⁻¹
Explanation:
1(a) We are given mass of product vs time. We need to find the concentration of reactant.
For example, at t = 0 min, there is 0 g of product. Therefore, there is 22.9 g − 0 g = 22.9 g of reactant. The molar mass is 60 g/mol, so there is (22.9 g) / (60 mol/g) = 0.382 mol. The volume is 1.0 L, so the concentration is (0.382 mol) / (1.0 L) = 0.382 M.
Next, we'll graph [A] vs time, ln[A] vs time, and 1/[A] vs time.
From the graphs, we can see that 1/[A] is linear. That means the reaction is second order.
1(b) The slope of the line is equal to the rate constant k.
k = 0.0596 M⁻¹min⁻¹
The half life is:
t = 1/(k [A₀])
where [A₀] is the initial concentration of reactant.
t = 1 / (0.0596 M⁻¹min⁻¹ × 0.382 M)
t = 44.0 min
(Notice the mass of product approximately doubles from t = 20 min to t = 45 min. This confirms that the half life is about 44 minutes.)
2(a) To make an Arrhenius graph, we need to graph ln(k) vs 1/T, where T is temperature in Kelvin. We'll make two graphs, one for ka (the forward reaction) and one for kb (the reverse reaction).
The slope of each line is -Eₐ/R, where R is the gas constant. For the forward reaction:
Eₐ = -8.314 × -10300 = 85,600 J/mol = 85.6 kJ/mol
For the reverse reaction:
Eₐ = -8.314 × -7227.6 = 60,100 J/mol = 60.1 kJ/mol
2(b) Make a graph showing the energy changes as the reaction goes from the reactants to the products. The difference between the reactants and the highest point is the forward activation energy. The difference between the products and the highest point is the reverse activation energy.
3) Like problem #1, we're going to graph [A], ln[A], and 1/[A] vs time.
Once again, 1/[A] vs t is linear, so this is a second order reaction.
The rate constant is the slope of this line, so k = 9.47 M⁻¹s⁻¹.
