What is bond enthalpy? Explain bond enthalpy in polyatomic molecules like $H_2O$.
(N/A) Bond enthalpy is defined as the amount of energy required to break one mole of bonds of a particular type between two atoms in the gaseous state.
The unit of bond enthalpy is $kJ \ mol^{-1}$.
For diatomic molecules,it is the bond dissociation enthalpy. For example,for $H_2$:
$H_{2(g)} \rightarrow H_{(g)} + H_{(g)} ; \Delta_{a} H^{\ominus} = 435.8 \ kJ \ mol^{-1}$
In polyatomic molecules like $H_2O$,the energy required to break the two $O-H$ bonds is not the same due to the change in the chemical environment after the first bond breaks:
$H_2O_{(g)} \rightarrow H_{(g)} + OH_{(g)} ; \Delta_{a} H_{1}^{\ominus} = 502 \ kJ \ mol^{-1}$
$OH_{(g)} \rightarrow H_{(g)} + O_{(g)} ; \Delta_{a} H_{2}^{\ominus} = 427 \ kJ \ mol^{-1}$
The average bond enthalpy for $O-H$ in water is the mean of these values: $(502 + 427) / 2 = 464.5 \ kJ \ mol^{-1}$.
This difference in energy values occurs because the electronic environment of the $O-H$ bond changes once one hydrogen atom is removed.
The unit of bond enthalpy is $kJ \ mol^{-1}$.
For diatomic molecules,it is the bond dissociation enthalpy. For example,for $H_2$:
$H_{2(g)} \rightarrow H_{(g)} + H_{(g)} ; \Delta_{a} H^{\ominus} = 435.8 \ kJ \ mol^{-1}$
In polyatomic molecules like $H_2O$,the energy required to break the two $O-H$ bonds is not the same due to the change in the chemical environment after the first bond breaks:
$H_2O_{(g)} \rightarrow H_{(g)} + OH_{(g)} ; \Delta_{a} H_{1}^{\ominus} = 502 \ kJ \ mol^{-1}$
$OH_{(g)} \rightarrow H_{(g)} + O_{(g)} ; \Delta_{a} H_{2}^{\ominus} = 427 \ kJ \ mol^{-1}$
The average bond enthalpy for $O-H$ in water is the mean of these values: $(502 + 427) / 2 = 464.5 \ kJ \ mol^{-1}$.
This difference in energy values occurs because the electronic environment of the $O-H$ bond changes once one hydrogen atom is removed.
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View SolutionFor the reaction,$2 H_{2(g)} + O_{2(g)} \longrightarrow 2 H_{2}O_{(g)}$,$\Delta H^{\circ} = -573.2 \ kJ$. The heat of decomposition of water per mole is
Find the value of $Q$ from the following equations:
$(i)$ $C_{(s)} + O_{2_{(g)}} \longrightarrow CO_{2_{(g)}}$ $\Delta H = Q \ kJ$
$(ii)$ $C_{(s)} + \frac{1}{2} O_{2_{(g)}} \longrightarrow CO_{(g)}$ $\Delta H = -x \ kJ$
$(iii)$ $CO_{(g)} + \frac{1}{2} O_{2_{(g)}} \longrightarrow CO_{2_{(g)}}$ $\Delta H = -y \ kJ$
$(i)$ $C_{(s)} + O_{2_{(g)}} \longrightarrow CO_{2_{(g)}}$ $\Delta H = Q \ kJ$
$(ii)$ $C_{(s)} + \frac{1}{2} O_{2_{(g)}} \longrightarrow CO_{(g)}$ $\Delta H = -x \ kJ$
$(iii)$ $CO_{(g)} + \frac{1}{2} O_{2_{(g)}} \longrightarrow CO_{2_{(g)}}$ $\Delta H = -y \ kJ$
The standard enthalpy of formation of $H_2O_{(g)}$ at $298 \ K$ is $-241.82 \ kJ \ mol^{-1}$. Calculate the enthalpy of formation of $H_2O_{(g)}$ at $373 \ K$,assuming $C_p$ is independent of temperature.
$C_p$ of $H_2O_{(g)} = 33.58 \ J \ K^{-1} \ mol^{-1}$
$C_p$ of $H_{2(g)} = 28.84 \ J \ K^{-1} \ mol^{-1}$
$C_p$ of $O_{2(g)} = 29.37 \ J \ K^{-1} \ mol^{-1}$
$C_p$ of $H_2O_{(g)} = 33.58 \ J \ K^{-1} \ mol^{-1}$
$C_p$ of $H_{2(g)} = 28.84 \ J \ K^{-1} \ mol^{-1}$
$C_p$ of $O_{2(g)} = 29.37 \ J \ K^{-1} \ mol^{-1}$
Medium
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