PCH241P OCT / NOV 2017 Question 2 (a) (i) Explain the difference between Exother
ID: 635430 • Letter: P
Question
PCH241P OCT / NOV 2017 Question 2 (a) (i) Explain the difference between Exothermic and Endothermic processes (u) Discuss 5 factors that can influence reaction rates (10) (b) Nitric oxide, NO, reacts with hydrogen as follows 2NO (g) H2 (g) N2O (g)H20 (g) Given the following data obtained during a senes of experiments, determine the reaction order of the reactants and the rate constant for the reaction Exp No [NO)/mol dm3 [H2 mol dm Rate/mol dm3 s 64 x 103 12 8 x 10 6 4 x 10 22 x 10 22 x 10 45 x 10 2 6 x 105 10 x 104 5 1 x 10 2 [20]Explanation / Answer
Part a
Exothermic process
An exothermic process in which heat releases by the system.
Reactants = products + heat
Endothermic process
An Endothermic process in which heat absorbed by the system.
Reactants + heat = products
Part b
5 factors influence reaction rates.
Temperature influences the reaction rate as the temperature increases, the number of collisions between reactants increase therefore reaction rate increases.
Surface area influences the reaction rate as surface area of particles increases, the frequency of collisions between reactants increase therefore reaction rate increases.
Pressure influences the reaction rate As the pressure of a gas increases, then the concentration also increases therefore reaction rate increases.
States influences the reaction rate the molecules of gaseous reactants have more kinetic energy so they react at a faster rate than solids. Rate of reaction of gasses reactants is faster than rate of reaction of solids.
Concentration influences the reaction rate as the concentration of reactant increases, the number of collisions between reactants increase therefore reaction rate increases.
Part c
Let rate law
r = k [NO]m [H2]n
From trial 1 and 2
r1/r2 = ([NO]1m [H2]1n) / ([NO]2m [H2]2n)
(2.6*10^-5)/(1*10^-4) = (6.4*10^3)m (2.2*10^-3)n/ (12.8*10^3)m (2.2*10^-3)n
0.26 = 0.5m
Take natural logarithm on both sides
ln 0.25 = m ln 0.5
m = 2
From trial 1 and 3
r1/r3 = ([NO]1m [H2]1n) / ([NO]3m [H2]3n)
(2.6*10^-5)/(5.1*10^-5) = (6.4*10^3)m (2.2*10^-3)n/ (6.4*10^3)m (4.5*10^-3)n
0.5 = 0.5n
n = 1
Rate law
r = k [NO]2 [H2]
Reaction order with respect to NO = 2
Reaction order with respect to H2 = 1
Overall order = 2 + 1 = 3
From trial 1
r1 = k [NO]12 [H2]1
2.6*10^-5 = k [6.4*10^3]2 [2.2*10^-3]
k = 2.88*10^-10 M-2 s-1
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