Question

For the reaction $\Delta H$

• A. $NaOH$
• B. $CH_{3}COOH$
• C. $CO(g)$
• D. None of these

Answer 1

Answer: (B)

$CH_{3}COOH$

Answer 2

$A \rightarrow B$

For this reaction $2C(s) + 2O_{2}(g) \rightarrow 2CO_{2}(g);\Delta H = - 787kJ$

$H_{2}(g) + \frac{1}{2}O_{2}(g) \rightarrow H_{2}O(l);\Delta H = - 286kJ$ is positive, i.e., there is an increase in the number of gaseous moles then $C_{2}H_{2}(g) + 2\frac{1}{2}O_{2}(g) \rightarrow 2CO_{2}(g) + H_{2}O(l);\Delta H = - 1301kJ$