JEE MAIN - Chemistry (2019 - 11th January Evening Slot - No. 9)
For the equilibrium,
2H2O $$\rightleftharpoons$$ H3O+ + OH$$-$$, the value of $$\Delta $$Go at 298 K is approximately :
2H2O $$\rightleftharpoons$$ H3O+ + OH$$-$$, the value of $$\Delta $$Go at 298 K is approximately :
$$-$$ 80 kJ molā1
100 kJ mol$$-$$1
$$-$$ 100 kJ mol$$-$$1
80 kJ molā1
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
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
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