1. Tardigrade
  2. Question
  3. Chemistry
  4. For the reaction CH4(g) + 2O2(g) arrow CO2(g) + 2H2O(l) Δ H°f of CH4(g), CO2(g) and H2O(l) is respectively -x, -y and -z kJ mol-1. Enthalpy change (∆r°H) for the reaction is

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Q. For the reaction $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$
$\Delta H^°_f\, of\, CH_4(g), CO_2(g)\, and\, H_2O(l)$ is respectively -x, -y and -z kJ $mol^{-1}$. Enthalpy change $(∆_r^°H)$ for the reaction is

Solution:

$\Delta H^{o}_{r}=\sum\Delta H_{f}^{o}$ = of products $-\sum\Delta H_{f}^{o}$ of reactants