1. At what temperature (in °C) does a reaction go from being spontaneous to nons
ID: 1042104 • Letter: 1
Question
1. At what temperature (in °C) does a reaction go from being spontaneous to nonspontaneous if it has ?H = -171 kJ/mol and ?S = -161 J/mol K ?
2. The Kb for a specific weak base has a value of 1.30 x 10-6, what is the pH of a 0.35 M aqueous solution of this weak base at 25 °C?
3. Consider the reaction below, for which ?G° = +1.2 kJ/mol at 25°C. Without knowing the pressure of the reactants and the product, we know that this reaction
X (g) + Y (g) <—> Z (g)
a. K = 0
b. K < 1
c. K = 1
d. K > 1
e. Can’t be determined
4. Which of the following can be behave as a lewis acid, and as a Bronsted-Lowery Acid?
H3O+ BF3 NH4+ Ca2+ HF H- Ne
please help with full explanation. thank you!
Explanation / Answer
1)
?Ho = -171.0 KJ/mol
?So = -161 J/mol.K
= -0.161 KJ/mol.K
use:
?Go = ?Ho - T*?So
for reaction to be spontaneous, ?Go should be negative
that is ?Go<0
since ?Go = ?Ho - T*?So
so, ?Ho - T*?So < 0
-171.0- T *-0.161 < 0
T *0.161 < 171.0
T < 1062 K
So, it is spontaneous for all temperature below 1062 K
which is (1062-273) oC = 789 oC
Answer: 789 oC
2)
B dissociates as:
B +H2O -----> BH+ + OH-
0.35 0 0
0.35-x x x
Kb = [BH+][OH-]/[B]
Kb = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Kb = x*x/(c)
so, x = sqrt (Kb*c)
x = sqrt ((1.3*10^-6)*0.35) = 6.745*10^-4
since c is much greater than x, our assumption is correct
so, x = 6.745*10^-4 M
So, [OH-] = x = 6.745*10^-4 M
use:
pOH = -log [OH-]
= -log (6.745*10^-4)
= 3.171
use:
PH = 14 - pOH
= 14 - 3.171
= 10.829
Answer: 10.83
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