Heat of combustion of two isomers x and y are | vedclass

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(D) The heat of combustion is a measure of the energy content of a compound. Higher heat of combustion indicates higher potential energy and lower sta

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isomer $x$ is $5 \ kJ/mol$ less stable

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(D) The heat of combustion is a measure of the energy content of a compound. Higher heat of combustion indicates higher potential energy and lower stability.Stability $\propto \frac{1}{	ext{Heat of combustion}}$.Given: Heat of combustion of $x = 17 \ kJ/mol$ and $y = 12 \ kJ/mol$.Since $x$ has a higher heat of combustion than $y$,$x$ is less stable than $y$.The difference in stability is $17 - 12 = 5 \ kJ/mol$.Therefore,isomer $x$ is $5 \ kJ/mol$ less stable than isomer $y$.

Heat of combustion of two isomers $x$ and $y$ are $17 \ kJ/mol$ and $12 \ kJ/mol$ respectively. From this information,it may be concluded that:

Similar Questions

The heat of combustion of a substance is:

$C(\text{diamond}) + O_{2(g)} \to CO_{2(g)}; \Delta H = -395 \text{ kJ}$
$C(\text{graphite}) + O_{2(g)} \to CO_{2(g)}; \Delta H = -393.5 \text{ kJ}$
If graphite is converted into diamond,then the $\Delta H$ for the process is . . . . . . $\text{kJ}$.

The enthalpy change for the transition of carbon from diamond to graphite is $\Delta H = -453.5 \ \text{cal}$. What does this indicate?

$H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}$; $\Delta H$ at $298 \ K = -285.8 \ kJ$. The molar enthalpy of vaporization of water at $1 \ atm$ and $25^{\circ}C$ is $44 \ kJ$. The standard enthalpy of formation of $1 \ mole$ of water vapor at $25^{\circ}C$ is $...... \ kJ$. (in $.8$)

The $\Delta H_f^{\circ}$ values (in $kJ \ mol^{-1}$) for graphite, diamond, and $C_{60}$ are respectively:

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