1
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 10th January Evening Slot

5.1 g NH4SH is introduced in 3.0 L evacuated flask at 327ºC. 30% of the solid NH4SH decomposed to NH3 and H2S as gases . The Kp of the reaction at 327oC is (R = 0.082 L atm mol–1 K–1, Molar mass of S = 32 g mol–1 molar mass of N = 14 g mol–1)
A
0.242 $$ \times $$ 10$$-$$4 atm2
B
1 $$ \times $$ 10–4 atm2
C
4.9 $$ \times $$ 10$$-$$3 atm2
D
0.242 atm2

Explanation

   NH4SH(s)  $$\rightleftharpoons$$    NH3(g) + H2S(g)

$$n = {{5.1} \over {51}} = .1\,mole\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,$$

$$.1\left( { - 1 - \alpha } \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.1\alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.1\alpha $$

$$\alpha \,\, = \,\,30\% = .3$$

so number of moles at equilibrium

$$\,\,\,\,\,\,\,.1\,(1 - .3)\,\,\,\,\,.1\,\, \times \,\,.3\,\,\,\,\,\,\,\,\,.1\,\, \times \,.3$$

$$ = \,\,\,\,.07\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = .03\;\,\,\,\,\,\,\,\,\, = .03$$

Now use PV = nRT at equilibrium

Ptotal $$ \times $$ 3 lit = (.03 + .03) $$ \times $$ .082 $$ \times $$ 600

Ptotal = .984 atm

At equilibrium

PNH3 = PH2S = $${{{P_{total}}} \over 2}$$ = .492

So kp = PNH3 . PH2S = (.492) (.492)

kp = .242 atm2
2
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 11th January Morning Slot

Consider the reaction
N2(g) + 3H2(g) $$\rightleftharpoons$$ 2NH3(g)

The equilibrium constant of the above reaction is Kp. If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by (Assume that PNH3 << Ptotal at equilibrium)
A
$${{{3^{{3 \over 2}}}{K_P^{{1 \over 2}}}{P^2}} \over 4}$$
B
$${{K_P^{{1 \over 2}}{P^2}} \over 4}$$
C
$${{{3^{{3 \over 2}}}{K_P^{{1 \over 2}}}{P^2}} \over 16}$$
D
$${{K_P^{{1 \over 2}}{P^2}} \over 16}$$

Explanation

N2(g) + 3H2(g) $$\rightleftharpoons$$ 2NH3(g) ; Keq = Kp

Write this equation reverse way,

2NH3(g)  $$\rightleftharpoons$$ N2(g) + 3H2(g) ; Keq = $${1 \over {{K_p}}}$$

2NH3(g) N2(g) + 3H2(g)
At t = 0 Po 0 0
At t = teq PNH3 p 3p

At equillibrium

PTotal = PNH3 + PN2 + PH2

= PNH3 + p + 3p

(As PNH3 << Ptotal so we can ignore PNH3)

$$ \therefore $$ PTotal = 4p

$$ \Rightarrow $$ p = $${{{{P_{total}}} \over 4}}$$

Formula of
Keq = $${{{p_{{N_2}}} \times {{\left( {{p_{{H_2}}}} \right)}^3}} \over {{{\left( {{p_{N{H_3}}}} \right)}^2}}}$$ = $${1 \over {{K_p}}}$$

$$ \Rightarrow $$ $${1 \over {{K_p}}}$$ = $${{p \times 27{p^3}} \over {{{\left( {{p_{N{H_3}}}} \right)}^2}}}$$

$$ \Rightarrow $$ $${{{\left( {{p_{N{H_3}}}} \right)}^2}}$$ = Kp $$ \times $$ 27 $$ \times $$ $${{{\left( {{{{P_{total}}} \over 4}} \right)}^4}}$$

$$ \Rightarrow $$ PNH3 = $$\sqrt {{K_p}} \times {\left( {27} \right)^{{1 \over 2}}} \times {\left( {{{{P_{Total}}} \over 4}} \right)^{{4 \over 2}}}$$

= $${{{3^{{3 \over 2}}}K_p^{{1 \over 2}}P_{Total}^2} \over {16}}$$
3
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 11th January Evening Slot

For the equilibrium,
2H2O $$\rightleftharpoons$$ H3O+ + OH$$-$$, the value of $$\Delta $$Go at 298 K is approximately :
A
$$-$$ 80 kJ mol–1
B
100 kJ mol$$-$$1
C
$$-$$ 100 kJ mol$$-$$1
D
80 kJ mol–1

Explanation

2H2O $$\rightleftharpoons$$ H3O+ + OH$$-$$

Here Keq = Kw(H2O)

At 298 K, Kw(H2O) = 10-14

$$ \therefore $$ Keq = 10-14

$$\Delta $$Go = -RTln(Keq)

= -RTln(Kw)

= -2.303RT log (10-14)

= -2.303 $$ \times $$ 8.314 $$ \times $$ 298 $$ \times $$ $$ \times $$ (-14)

= 80 kJ/mol
4
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 11th January Evening Slot

Given the equilibrium constant:

KC of the reaction :

Cu(s) + 2Ag+ (aq) $$ \to $$ Cu2+ (aq) + 2Ag(s) is

10 $$ \times $$ 1015, calculate the E$$_{cell}^0$$ of this reaciton at 298 K

[2.303 $${{RT} \over F}$$ at 298 K = 0.059V]
A
0.4736 mV
B
0.04736 V
C
0.4736 V
D
0.04736 mV

Explanation

We know,

$$\Delta $$Go = -RTln(KC) ....(1)

Also $$\Delta $$Go = -nF$$E_{cell}^o$$ ....(2)

$$ \therefore $$ -nF$$E_{cell}^o$$ = -RTln(KC)

$$ \Rightarrow $$ $$E_{cell}^o$$ = $${{RT} \over {nF}}\ln \left( {{K_C}} \right)$$

= $$2.303{{RT} \over {nF}}\log \left( {{K_C}} \right)$$

= $${{0.059} \over 2}\log \left( {10 \times {{10}^{15}}} \right)$$

( n = no of electron transferred = 2 )

= 0.059 $$ \times $$ $${{16} \over 2}$$

= 0.059 $$ \times $$ 8

= 0.472 V

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