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If $2 \mathrm{~kJ}$ of heat is released from system and $6 \mathrm{~kJ}$ of work is done on the system, what is enthalpy change of system?
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The correct answer is:
$-2 \mathrm{k} \mathrm{J}$
(A)
$Q=-2 \mathrm{~kJ}, \quad W=+6 \mathrm{~kJ}$ $\therefore \quad \Delta U=Q+W=-2+6=4 \mathrm{~kJ}$
Now, $\Delta H=\Delta U-W=4-6=-2 k J$
$Q=-2 \mathrm{~kJ}, \quad W=+6 \mathrm{~kJ}$ $\therefore \quad \Delta U=Q+W=-2+6=4 \mathrm{~kJ}$
Now, $\Delta H=\Delta U-W=4-6=-2 k J$
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