JEE MAIN - Chemistry (2016 - 9th April Morning Slot - No. 19)
For the reaction,
A(g) + B(g) $$ \to $$ C(g) + D(g), $$\Delta $$Ho and $$\Delta $$So are, respectively, − 29.8 kJ mol−1 and −0.100 kJ K−1 mol−1 at 298 K. The equilibrium constant for the reaction at 298 K is :
A(g) + B(g) $$ \to $$ C(g) + D(g), $$\Delta $$Ho and $$\Delta $$So are, respectively, − 29.8 kJ mol−1 and −0.100 kJ K−1 mol−1 at 298 K. The equilibrium constant for the reaction at 298 K is :
1.0 $$ \times $$ 10$$-$$10
1.0 $$ \times $$ 1010
10
1
Explanation
We have
$$\Delta$$G$$^\circ$$ = $$\Delta$$H$$^\circ$$ $$-$$ T$$\Delta$$S$$^\circ$$
$$\Delta$$G$$^\circ$$ = $$-$$ 29.8 $$-$$ 298 ($$-$$ 0.1) = 0
$$\Delta$$G$$^\circ$$ = $$-$$ 2.302 RT log Keq
log Keq = 0 $$\Rightarrow$$ Keq = 1
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