If the value of equilibrium constant for a particular reaction is 1.6 $$ \times $$ 1012, then at equilibrium the system will contain
A
mostly products
B
similar amounts of reactants and products
C
all reactants
D
mostly reactants.
Explanation
As, 1.6 × 1012 is very high value of K. Thus,
the reaction proceeds almost to completion and mixture must contain mostly products.
3
AIPMT 2015 Cancelled Paper
MCQ (Single Correct Answer)
Which of the following statements is correct for a reversible process in a state of equilibrium?
A
$$\Delta $$Go = $$-$$2.30 RT log K
B
$$\Delta $$Go = 2.30 RT log K
C
$$\Delta $$G = $$-$$2.30 RT log K
D
$$\Delta $$G = 2.30 RT log K
Explanation
$$\Delta $$G = $$\Delta $$Go + 2.303 RT log Q
At equilibrium, when $$\Delta $$G = 0 and Q = K
then 0 = $$\Delta $$Go + 2.303 RT log K
$$\Delta $$Go = – 2.303 RT log K
4
AIPMT 2015 Cancelled Paper
MCQ (Single Correct Answer)
The Ksp of Ag2CrO4, AgCl, AgBr and Agl are respectively, 1.1 $$ \times $$ 10$$-$$12, 1.8 $$ \times $$ 10$$-$$10, 5.0 $$ \times $$ 10$$-$$13, 8.3 $$ \times $$ 10$$-$$17. Which one of the following salts will precipitate last if AgNO3 solution is added to the solution containing equal moles of NaCl, NaBr, Nal and Na2CrO4?
A
AgBr
B
Ag2CrO4
C
Agl
D
AgCl
Explanation
From the Ksp values of the given salts calculate
the solubility values. Salt having highest solubility
will precipitate at last.
AgCrO4
⇌
2Ag+
+
CrO42-
s
2s
s
Ksp = (2s)2(s) = 1.1 × 10–12
$$ \Rightarrow $$ s = 0.65 × 10–4
AgCl
⇌
Ag+
+
Cl-
s
s
s
Ksp = s × s
$$ \Rightarrow $$ 1.8 × 10–10 = s
2
$$ \Rightarrow $$ s = 1.34 × 10–5
AgBr
⇌
Ag+
+
Br-
s
s
s
Ksp = s × s
$$ \Rightarrow $$ 5 × 10–13 = s2
$$ \Rightarrow $$ s = 0.71 × 10–6
AgI
⇌
Ag+
+
I-
s
s
s
Ksp = s × s
$$ \Rightarrow $$ 8.3 × 10–17 = s2
$$ \Rightarrow $$ s = 0.9 × 10–8
$$ \therefore $$ Solubility of Ag2CrO4 is maximum so, it
will precipitate at last.
Questions Asked from Equilibrium
On those following papers in MCQ (Single Correct Answer)
Number in Brackets after Paper Indicates No. of Questions
