JEE MAIN - Chemistry (2017 - 8th April Morning Slot - No. 23)
For a reaction, A(g) $$ \to $$ A($$\ell $$); $$\Delta $$H= $$-$$ 3RT.
The correct statement for the reaction is :
The correct statement for the reaction is :
$$\Delta $$H = $$\Delta $$U $$ \ne $$ O
$$\Delta $$H = $$\Delta $$U = O
$$\left| {} \right.$$$$\Delta $$H$$\left| {} \right.$$ < $$\left| {} \right.$$$$\Delta $$U$$\left| {} \right.$$
$$\left| {} \right.$$$$\Delta $$H$$\left| {} \right.$$ > $$\left| {} \right.$$$$\Delta $$U$$\left| {} \right.$$
Explanation
Given reaction,
A(g) $$ \to $$ A($$l$$) ;
We know,
$$\Delta $$H = $$\Delta $$V + $$\Delta $$ng RT
Given, $$\Delta $$H = $$-$$3RT
From reaction,
$$\Delta $$ng = 0$$-$$ 1 = $$-1$$
$$\therefore\,\,\,$$ $$\Delta $$H = $$\Delta $$U $$-$$ RT
$$ \Rightarrow $$ $$\,\,\,$$ $$-$$ 3RT = $$\Delta $$U $$-$$ RT
$$ \Rightarrow $$$$\,\,\,$$ $$\Delta $$U = $$-$$ 2RT
$$\therefore\,\,\,$$ $$\left| {\Delta U} \right|$$ = 2RT
and $$\left| {\Delta H} \right|$$ = 3RT
$$\therefore\,\,\,$$ $$\left| {\Delta H} \right|$$ > $$\left| {\Delta U} \right|$$
A(g) $$ \to $$ A($$l$$) ;
We know,
$$\Delta $$H = $$\Delta $$V + $$\Delta $$ng RT
Given, $$\Delta $$H = $$-$$3RT
From reaction,
$$\Delta $$ng = 0$$-$$ 1 = $$-1$$
$$\therefore\,\,\,$$ $$\Delta $$H = $$\Delta $$U $$-$$ RT
$$ \Rightarrow $$ $$\,\,\,$$ $$-$$ 3RT = $$\Delta $$U $$-$$ RT
$$ \Rightarrow $$$$\,\,\,$$ $$\Delta $$U = $$-$$ 2RT
$$\therefore\,\,\,$$ $$\left| {\Delta U} \right|$$ = 2RT
and $$\left| {\Delta H} \right|$$ = 3RT
$$\therefore\,\,\,$$ $$\left| {\Delta H} \right|$$ > $$\left| {\Delta U} \right|$$
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