1
MCQ (Single Correct Answer)

JEE Main 2018 (Online) 15th April Evening Slot

$$\Delta $$fGo at 500 K for substance 'S' in liquid state and gaseous state are +100.7 kcl mol-1 and +103 kcal mol-1, respectively. Vapour pressure of liquid 'S' at 500 K is approximately equal to : ( R = 2 cal K-1 mol-1 )
A
0.1 atm
B
1 atm
C
10 atm
D
100 atm

Explanation

S($$l$$)   $$\buildrel \, \over \longrightarrow $$   S(g)

$$\Delta $$Go = $$\Delta $$fGo (Vapour) $$-$$ $$\Delta $$f Go (liquid)

= 103 $$-$$ 100.7

= 2.3 kcal / mol

= 2.3 $$ \times $$ 103 cal/mol.

$$\Delta $$Go = $$-$$ RT$$l$$nk

$$ \Rightarrow $$$$\,\,\,$$ 2.3 $$ \times $$ 103 = $$-$$ 2.303 $$ \times $$ 2 $$ \times $$ 500 log K

$$ \Rightarrow $$$$\,\,\,$$ log K = $$-$$ 1

$$ \Rightarrow $$$$\,\,\,$$ K = 0.1 atm
2
MCQ (Single Correct Answer)

JEE Main 2018 (Online) 16th April Morning Slot

At 320 K, a gas A2 is 20% dissociated to A(g). The standard free energy change at 320 K and 1 atm in J mol-1 is approximately : (R = 8.314 JK-1 mol-1; ln2 = 0.693; ln 3 = 1.098)
A
4763
B
2068
C
1844
D
4281

Explanation

A2 (g) $$\rightleftharpoons$$ 2 A (g)

Assume initially concentration of A2 = [A2] = 1 m

at equilibrium [A2] = 1 $$ \times $$ $${{80} \over {100}}$$ = 0.8 M

and 20% of [A2] = 1 $$ \times $$ $${{20} \over {100}}$$ = 0.2 M

$$\therefore\,\,\,\,$$ [A] = 2 $$ \times $$ 0.2 = 0.4 M

Equilibrium constant

K = $${{{{[A]}^2}} \over {[{A_2}]}}$$ = $${{{{\left[ {0.4} \right]}^2}} \over {\left[ {0.8} \right]}}$$ = 0.2

$$\Delta $$Go = $$-$$ RT $$\ell $$nK

= $$-$$ 8.314 $$ \times $$ 320 $$ \times $$ $$\ell $$n(0.2)

= 4281 J/mol.
3
MCQ (Single Correct Answer)

JEE Main 2018 (Online) 16th April Morning Slot

For which of the following processes, $$\Delta $$S is negative ?
A
H2(g) $$ \to $$ 2H(g)
B
N2(g, 1 atm) $$ \to $$ N2(g, 5 atm)
C
C(diamond) $$ \to $$ C(graphite)
D
N2(g, 273 K) $$ \to $$ N2(g, 300 K)

Explanation

N2(g, 1 atom) $$\buildrel \, \over \longrightarrow $$ N2(g, 5 atom)

Here pressure increases. When pressure increases then the molecules of will come closer and intermoleculer distance decreases, so entropy will also decreases and $$\Delta S\, < \,0$$.
4
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Morning Slot

Which one of the following statements regarding Henry's law is not correct ?
A
Higher the value of KH at a given pressure, higher is the solubility of the gas in the liquids
B
Different gases have different KH (Henry's law constant) values at the same temperature.
C
The partial pressure of the gas in vapour phase is proportional to the mole fraction of the gas in the solution.
D
The value of KH increases with increase of temperature and KH is function of the nature of the gas

Explanation

From Henry's law we know,

Pgas = KH xg

Where, Pgas = Pressure of undissolved gas

xg = Mole fraction of gas dissolved into the liquid.

KH = Henry's Carnot.

When pressure is constant then,

xg $$ \propto $$ $${1 \over {K{}_H}}$$

So. when KH is high then xg or solubility of gas is lower in the liquid.

So, option (A) is wrong.

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