Calculate the enthalpy change for the process $CCl_{4(g)} \to C_{(g)} + 4Cl_{(g)}$ and calculate the bond enthalpy of the $C-Cl$ bond in $CCl_{4(g)}$.
$\Delta_{vap} H^{\theta}(CCl_{4}) = 30.5 \, kJ \, mol^{-1}$
$\Delta_{f} H^{\theta}(CCl_{4}) = -135.5 \, kJ \, mol^{-1}$
$\Delta_{a} H^{\theta}(C) = 715.0 \, kJ \, mol^{-1}$ (where $\Delta_{a} H^{\theta}$ is enthalpy of atomisation)
$\Delta_{a} H^{\theta}(Cl_{2}) = 242 \, kJ \, mol^{-1}$
$\Delta_{vap} H^{\theta}(CCl_{4}) = 30.5 \, kJ \, mol^{-1}$
$\Delta_{f} H^{\theta}(CCl_{4}) = -135.5 \, kJ \, mol^{-1}$
$\Delta_{a} H^{\theta}(C) = 715.0 \, kJ \, mol^{-1}$ (where $\Delta_{a} H^{\theta}$ is enthalpy of atomisation)
$\Delta_{a} H^{\theta}(Cl_{2}) = 242 \, kJ \, mol^{-1}$
The chemical equations for the given enthalpy values are:
$(i) \quad CCl_{4(l)} \longrightarrow CCl_{4(g)} \quad \Delta_{vap}H^{\theta} = 30.5 \, kJ \, mol^{-1}$
$(ii) \quad C_{(s)} \longrightarrow C_{(g)} \quad \Delta_{a}H^{\theta} = 715.0 \, kJ \, mol^{-1}$
$(iii) \quad Cl_{2(g)} \longrightarrow 2Cl_{(g)} \quad \Delta_{a}H^{\theta} = 242 \, kJ \, mol^{-1}$
$(iv) \quad C_{(s)} + 2Cl_{2(g)} \longrightarrow CCl_{4(l)} \quad \Delta_{f}H^{\theta} = -135.5 \, kJ \, mol^{-1}$
To find the enthalpy change for $CCl_{4(g)} \longrightarrow C_{(g)} + 4Cl_{(g)}$,we rearrange the equations:
$\Delta H = \Delta_{a}H^{\theta}(C) + 2\Delta_{a}H^{\theta}(Cl_{2}) - \Delta_{vap}H^{\theta} - \Delta_{f}H^{\theta}$
$\Delta H = 715.0 + 2(242) - 30.5 - (-135.5)$
$\Delta H = 715.0 + 484 - 30.5 + 135.5 = 1304 \, kJ \, mol^{-1}$
Bond enthalpy of $C-Cl$ bond in $CCl_{4(g)} = \frac{1304}{4} = 326 \, kJ \, mol^{-1}$.
$(i) \quad CCl_{4(l)} \longrightarrow CCl_{4(g)} \quad \Delta_{vap}H^{\theta} = 30.5 \, kJ \, mol^{-1}$
$(ii) \quad C_{(s)} \longrightarrow C_{(g)} \quad \Delta_{a}H^{\theta} = 715.0 \, kJ \, mol^{-1}$
$(iii) \quad Cl_{2(g)} \longrightarrow 2Cl_{(g)} \quad \Delta_{a}H^{\theta} = 242 \, kJ \, mol^{-1}$
$(iv) \quad C_{(s)} + 2Cl_{2(g)} \longrightarrow CCl_{4(l)} \quad \Delta_{f}H^{\theta} = -135.5 \, kJ \, mol^{-1}$
To find the enthalpy change for $CCl_{4(g)} \longrightarrow C_{(g)} + 4Cl_{(g)}$,we rearrange the equations:
$\Delta H = \Delta_{a}H^{\theta}(C) + 2\Delta_{a}H^{\theta}(Cl_{2}) - \Delta_{vap}H^{\theta} - \Delta_{f}H^{\theta}$
$\Delta H = 715.0 + 2(242) - 30.5 - (-135.5)$
$\Delta H = 715.0 + 484 - 30.5 + 135.5 = 1304 \, kJ \, mol^{-1}$
Bond enthalpy of $C-Cl$ bond in $CCl_{4(g)} = \frac{1304}{4} = 326 \, kJ \, mol^{-1}$.
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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 .............. $\text{kJ mol}^{-1}$
| $H-H$ 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 .............. $\text{kJ mol}^{-1}$
MediumAIPMT 2009
View SolutionFill in the blanks:
$(i)$ The enthalpy change that occurs when one mole of a compound is formed from its constituent elements in their standard states is called the ...... of that compound.
$(ii)$ The total heat change in a chemical reaction is equal to the algebraic sum of the heat changes of the individual steps of the reaction. This law was given by ............. .
$(iii)$ $A$ process during which there is no exchange of heat between the system and the surroundings is called an ................ process.
$(iv)$ The standard enthalpy value of any element in its standard state is considered to be ...... .
$(i)$ The enthalpy change that occurs when one mole of a compound is formed from its constituent elements in their standard states is called the ...... of that compound.
$(ii)$ The total heat change in a chemical reaction is equal to the algebraic sum of the heat changes of the individual steps of the reaction. This law was given by ............. .
$(iii)$ $A$ process during which there is no exchange of heat between the system and the surroundings is called an ................ process.
$(iv)$ The standard enthalpy value of any element in its standard state is considered to be ...... .
Difficult
View SolutionGiven that bond energies of $H-H$ and $Cl-Cl$ are $430 \ kJ \ mol^{-1}$ and $240 \ kJ \ mol^{-1}$ respectively and $\Delta H_f$ for $HCl$ is $-90 \ kJ \ mol^{-1},$ the bond enthalpy of $HCl$ is ............... $kJ \ mol^{-1}$.
MediumAIPMT 2007
View SolutionWhich of the following compounds will absorb the maximum quantity of heat when dissolved in the same amount of water? The heats of solution of these compounds at $25 \, ^\circ C$ in $kJ/mole$ of each solute are given in brackets.
Medium
View SolutionFrom the following data at $25^{\circ} C$,calculate the $\Delta_{r} H^0$ for the reaction $H_2O_{(g)} \rightarrow 2 H_{(g)} + O_{(g)}$:
$1/2 H_{2(g)} + 1/2 O_{2(g)} \rightarrow OH_{(g)}$ $\Delta H = 42.09 \ kJ \ mol^{-1}$ $H_{2(g)} + 1/2 O_{2(g)} \rightarrow H_2O_{(g)}$ $\Delta H = -242 \ kJ \ mol^{-1}$ $H_{2(g)} \rightarrow 2 H_{(g)}$ $\Delta H = 436 \ kJ \ mol^{-1}$ $O_{2(g)} \rightarrow 2 O_{(g)}$ $\Delta H = 496 \ kJ \ mol^{-1}$
| $1/2 H_{2(g)} + 1/2 O_{2(g)} \rightarrow OH_{(g)}$ | $\Delta H = 42.09 \ kJ \ mol^{-1}$ |
| $H_{2(g)} + 1/2 O_{2(g)} \rightarrow H_2O_{(g)}$ | $\Delta H = -242 \ kJ \ mol^{-1}$ |
| $H_{2(g)} \rightarrow 2 H_{(g)}$ | $\Delta H = 436 \ kJ \ mol^{-1}$ |
| $O_{2(g)} \rightarrow 2 O_{(g)}$ | $\Delta H = 496 \ kJ \ mol^{-1}$ |
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