
Calculate $\Delta {H^0}r$ for the reaction $C{H_2}C{l_2}(g) \to C(g) + 2H(g) + 2Cl(g)$. The average bond enthalpies of C-H and C-Cl bonds are $414 KJmol^-1$ and $330 KJmol^-1$.
Answer
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Hint:Bond enthalpy is the measure of the strength of the bond. It is also known as the mean bond enthalpy. It is given by the IUPAC as the measure of the bond dissociation enthalpy in the gaseous state of the same element.
Complete step by step answer:
The bond enthalpy is usually calculated for the one mole amount of the substance. The standard energy required for the formation of the compounds from its constituent atoms and molecules in their stable state is known as the enthalpy of formation and is calculated at 1 bar pressure and 298.15 k under the standard conditions of temperature and pressure.
Certain assumptions are also used during the calculation:
It is assumed that the gas obeys the ideal gas equation.
The concentration of the solution is assumed to be one mole per litre.
The physical states of the substance are considered under the standard conditions of the pressure.
The physical states of the reaction are always mentioned along with the balanced chemical equations during the calculation of the standard enthalpy of formation.
Given in the question: Average bond enthalpies of C-H and C-Cl bonds are $414 KJmol^-1$ and $330 KJmol^-1$
From the reaction: $C{H_2}C{l_2}(g) \to C(g) + 2H(g) + 2Cl(g)$
In the formation of the $C{H_2}C{l_2}(g)$ molecule the bond enthalpies of the C-H and C-Cl bond plays the major role and thus the expression can be represented in the mathematical form as:
It is evident that:
$ \Delta {H_0}^r = 2\Delta {H_{(C - H)}} + 2\Delta {H_{(C - Cl)}} \\
= 2 \times 414 + 2 \times 330 \\
= 1488KJmo{l^{ - 1}} \\ $
Thus the $\Delta {H^0}r$ for the reaction $C{H_2}C{l_2}(g) \to C(g) + 2H(g) + 2Cl(g)$ is 1488 KJ/ mol.
Note:
The formation constant derived during the formation of the particular reactant from its constituent atoms or molecules, is a dimensionless quantity and remains as a constant obtained under the standard conditions of temperature and pressure.
Complete step by step answer:
The bond enthalpy is usually calculated for the one mole amount of the substance. The standard energy required for the formation of the compounds from its constituent atoms and molecules in their stable state is known as the enthalpy of formation and is calculated at 1 bar pressure and 298.15 k under the standard conditions of temperature and pressure.
Certain assumptions are also used during the calculation:
It is assumed that the gas obeys the ideal gas equation.
The concentration of the solution is assumed to be one mole per litre.
The physical states of the substance are considered under the standard conditions of the pressure.
The physical states of the reaction are always mentioned along with the balanced chemical equations during the calculation of the standard enthalpy of formation.
Given in the question: Average bond enthalpies of C-H and C-Cl bonds are $414 KJmol^-1$ and $330 KJmol^-1$
From the reaction: $C{H_2}C{l_2}(g) \to C(g) + 2H(g) + 2Cl(g)$
In the formation of the $C{H_2}C{l_2}(g)$ molecule the bond enthalpies of the C-H and C-Cl bond plays the major role and thus the expression can be represented in the mathematical form as:
It is evident that:
$ \Delta {H_0}^r = 2\Delta {H_{(C - H)}} + 2\Delta {H_{(C - Cl)}} \\
= 2 \times 414 + 2 \times 330 \\
= 1488KJmo{l^{ - 1}} \\ $
Thus the $\Delta {H^0}r$ for the reaction $C{H_2}C{l_2}(g) \to C(g) + 2H(g) + 2Cl(g)$ is 1488 KJ/ mol.
Note:
The formation constant derived during the formation of the particular reactant from its constituent atoms or molecules, is a dimensionless quantity and remains as a constant obtained under the standard conditions of temperature and pressure.
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