Given,
$NO_{(g)} + O_{3(g)} \longrightarrow NO_{2(g)} + O_{2(g)}; \Delta H = -198.9 \, kJ/mol$
$O_{3(g)} \longrightarrow 3/2 O_{2(g)}; \Delta H = -142.3 \, kJ/mol$
$O_{2(g)} \longrightarrow 2O_{(g)}; \Delta H = +495.0 \, kJ/mol$
The enthalpy change $(\Delta H)$ for the following reaction is $..... \, kJ/mol$
$NO_{(g)} + O_{(g)} \longrightarrow NO_{2(g)}$
$NO_{(g)} + O_{3(g)} \longrightarrow NO_{2(g)} + O_{2(g)}; \Delta H = -198.9 \, kJ/mol$
$O_{3(g)} \longrightarrow 3/2 O_{2(g)}; \Delta H = -142.3 \, kJ/mol$
$O_{2(g)} \longrightarrow 2O_{(g)}; \Delta H = +495.0 \, kJ/mol$
The enthalpy change $(\Delta H)$ for the following reaction is $..... \, kJ/mol$
$NO_{(g)} + O_{(g)} \longrightarrow NO_{2(g)}$
- A$-304.1$
- B$+304.1$
- C$-403.1$
- D$+403.1$
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View SolutionGiven:
$I$. $2Fe_{(s)} + \frac{3}{2} O_{2(g)} \to Fe_2O_{3(s)}$; $\Delta H^{\Theta} = -193.4 \ kJ$
$II$. $Mg_{(s)} + \frac{1}{2} O_{2(g)} \to MgO_{(s)}$; $\Delta H^{\Theta} = -140.2 \ kJ$
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....... $kJ$
$I$. $2Fe_{(s)} + \frac{3}{2} O_{2(g)} \to Fe_2O_{3(s)}$; $\Delta H^{\Theta} = -193.4 \ kJ$
$II$. $Mg_{(s)} + \frac{1}{2} O_{2(g)} \to MgO_{(s)}$; $\Delta H^{\Theta} = -140.2 \ kJ$
What is $\Delta H^{\Theta}$ of the reaction?
$3Mg_{(s)} + Fe_2O_{3(s)} \to 3MgO_{(s)} + 2Fe_{(s)}$
....... $kJ$
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