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:

  • A
    isomer $x$ is $5 \ kJ/mol$ more stable
  • B
    isomer $y$ is $5 \ kJ/mol$ less stable
  • C
    isomer $y$ has $5 \ kJ/mol$ more potential energy
  • D
    isomer $x$ is $5 \ kJ/mol$ less stable

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Similar Questions

Which of the following is an example of an endothermic reaction?

Based on the following thermochemical equations:
$H_2O_{(g)} + C_{(s)} \to CO_{(g)} + H_{2(g)}; \Delta H = 131 \ kJ$
$CO_{(g)} + \frac{1}{2}O_{2(g)} \to CO_{2(g)}; \Delta H = -282 \ kJ$
$H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(g)}; \Delta H = -242 \ kJ$
$C_{(s)} + O_{2(g)} \to CO_{2(g)}; \Delta H = X \ kJ$
The value of $X$ is ...... $kJ$.

The combustion enthalpies of carbon,hydrogen,and methane are $-393.5 \ kJ/mole$,$-284.8 \ kJ/mole$,and $-890.4 \ kJ/mole$ respectively at $25 \ ^oC$. The value of the standard formation enthalpy of methane at that temperature is ..... $kJ/mole$.

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For the following reaction,
$C (diamond) + O_2 \rightarrow CO_{2(g)}$; $\Delta H = -97.6 \ kcal$
$C (graphite) + O_2 \rightarrow CO_{2(g)}$; $\Delta H = -94.3 \ kcal$
The heat change for the conversion of $1 \ g$ of $C (diamond) \rightarrow C (graphite)$ is (in $kcal$)

From the following thermochemical equations,find out the heat of ionisation $Z$ for a weak acid $HA$:
$HA \to A^{-} + H^{+} , \Delta H = Z$
$H^{+} + OH^{-} \to H_2O , \Delta H = X$
$HA + OH^{-} \to A^{-} + H_2O , \Delta H = Y$

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