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 Mā»Ā¹sā»Ā¹
Explanation:
1(a) We are given mass of product vs time. Ā We need to find the concentration of reactant.
For example, at t = 0 min, there is 0 g of product. Ā Therefore, there is 22.9 g ā 0 g = 22.9 g of reactant. Ā The molar mass is 60 g/mol, so there is (22.9 g) / (60 mol/g) = 0.382 mol. Ā 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. Ā That means the reaction is second order.
1(b) The slope of the line is equal to the rate constant k.
k = 0.0596 Mā»Ā¹minā»Ā¹
The half life is:
t = 1/(k [Aā])
where [Aā] is the initial concentration of reactant.
t = 1 / (0.0596 Mā»Ā¹minā»Ā¹ Ć 0.382 M)
t = 44.0 min
(Notice the mass of product approximately doubles from t = 20 min to t = 45 min. Ā 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. Ā We'll make two graphs, one for ka (the forward reaction) and one for kb (the reverse reaction).
The slope of each line is -Eā/R, where R is the gas constant. Ā For the forward reaction:
Eā = -8.314 Ć -10300 = 85,600 J/mol = 85.6 kJ/mol
For the reverse reaction:
Eā = -8.314 Ć -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. Ā The difference between the reactants and the highest point is the forward activation energy. Ā 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 Mā»Ā¹sā»Ā¹.
