JEE MAIN - Chemistry (2018 - 15th April Evening Slot - No. 13)
$$\Delta $$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 )
0.1 atm
1 atm
10 atm
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
$$\Delta $$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
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