JEE MAIN - Chemistry (2019 - 10th January Evening Slot - No. 12)
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)
0.242 $$ \times $$ 10$$-$$4 atm2
1 $$ \times $$ 10–4 atm2
4.9 $$ \times $$ 10$$-$$3 atm2
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
$$n = {{5.1} \over {51}} = .1\,mole\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,$$
$$.1\left( { - 1 - \alpha } \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.1\alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.1\alpha $$
$$\alpha \,\, = \,\,30\% = .3$$
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
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