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 $..... \, kJ/mol$
$NO_{(g)} + O_{(g)} \longrightarrow NO_{2(g)}$

  • A
    $-304.1$
  • B
    $+304.1$
  • C
    $-403.1$
  • D
    $+403.1$

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Similar Questions

According to Hess's Law,the enthalpy change of a reaction depends on which of the following?

Choose the reaction$(s)$ from the following options,for which the standard enthalpy of reaction is equal to the standard enthalpy of formation.
$(1)$ $\frac{3}{2} O_{2(g)} \rightarrow O_{3(g)}$
$(2)$ $\frac{1}{8} S_{8(s)} + O_{2(g)} \rightarrow SO_{2(g)}$
$(3)$ $2 H_{2(g)} + O_{2(g)} \rightarrow 2 H_2O_{(l)}$
$(4)$ $2 C_{(g)} + 3 H_{2(g)} \rightarrow C_2H_{6(g)}$

The heat of neutralisation of one equivalent of an acid by one equivalent of a base is minimum when:

$18.0 \ g$ of water completely vaporises at $100^{\circ}C$ and $1 \ bar$ pressure and the enthalpy change in the process is $40.79 \ kJ \ mol^{-1}$. What will be the enthalpy change for vaporising two moles of water under the same conditions? What is the standard enthalpy of vaporisation for water?

From the following bond energies:
$H-H$ bond energy$431.37 \text{ kJ mol}^{-1}$
$C=C$ bond energy$606.10 \text{ kJ mol}^{-1}$
$C-C$ bond energy$336.49 \text{ kJ mol}^{-1}$
$C-H$ bond energy$410.50 \text{ kJ mol}^{-1}$

Enthalpy for the reaction $CH_2=CH_2 + H-H \to CH_3-CH_3$ will be .............. $\text{kJ mol}^{-1}$

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