Question

The relation between change in internal energy $HNO_{3}$, change in enthalpy $NaOH$ and work done (W) is represented as.

• A. $HCl$
• B. $C_{6}H_{6} + 7\frac{1}{2}O_{2} \rightarrow 6CO_{2}(g) + 3H_{2}O(l);\Delta H = - 3264.6kJ$
• C. $39gC_{6}H_{6}$
• D. $C(\text{graphite}) + \frac{1}{2}O_{2}(g) \rightarrow CO(g);\Delta H = - 110.5kJ$

Answer 1

Answer: (A)

$HCl$

Answer 2

$\Delta H = T\Delta S$.