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
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