1
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Evening Slot

The pH of rain water, is approximately :
A
5.6
B
7.5
C
7.0
D
6.5

Explanation

pH of rain water is approximate 5.6
2
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Evening Slot

If the standard electrode potential for a cell is 2 V at 300 K, the equilibrium constant (K) for the reaction
Zn(s) + Cu2+ (aq) $$\rightleftharpoons$$ Zn2+(aq) + Cu(s)
at 300 K is approximately
(R = 8 JK$$-$$1mol$$-$$1, F = 96000 C mol$$-$$1)
A
e$$-$$80
B
e$$-$$160
C
e320
D
e160

Explanation

$$\Delta $$Go = $$-$$ RT lnk = $$-$$nFEocell

lnk = $${{n \times F \times {E^o}} \over {R \times T}} = {{2 \times 96000 \times 2} \over {8 \times 300}}$$

lnk = 160

k = e160
3
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
4
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 = $$\sqrt {{K_p}} \times {\left( {27} \right)^{{1 \over 2}}} \times {\left( {{{{P_{Total}}} \over 4}} \right)^{{4 \over 2}}}$$

= $${{{3^{{3 \over 2}}}K_p^{{1 \over 2}}P_{Total}^2} \over {16}}$$

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