The bond dissociation energy of gaseous ${H_2}$ , $C{l_2}$ and $HCl$ are 104, 58 and 103 $kcal\,mo{l^{ - 1}}$ respectively. The enthalpy formation for $HCl$ gas will be:
A. -44.0 $kcal$
B. -22.0 $kcal$
C. 22.0 $kcal$
D. 44.0 $kcal$
Answer
630.6k+ views
Hint: In order to the question, first we will discuss about the enthalpy formation then we should write the exact balanced reaction according to the given gaseous compounds and according to the reactants and product, we will find the enthalpy formation of $HCl$ compound.
Complete step-by-step solution:Bond dissociation energy is the energy required to break a chemical bond. It is one method for measuring the strength of a compound bond. Bond separation energy approaches bond energy just for diatomic atoms. The most grounded bond separation energy is for the Si-F bond. The most fragile energy is for a covalent bond and is similar.
The enthalpy of formation is the standard reaction enthalpy for the formation of the compound from its elements (atoms or molecules) in their most stable reference states at the chosen temperature (298.15K) and at 1bar pressure.
First, we’ll write the balanced reaction between the given compounds ${H_2}$ , $C{l_2}$ and $HCl$
$\dfrac{1}{2}{H_2} + \dfrac{1}{2}C{l_2} \to HCl$
According to the above reaction, we have the formula to find the enthalpy formation for the product $HCl$ gas:
$\Delta {h_f} = (Bond\,energy\,of\,reac\tan t) - (Bond\,energy\,of\,\Pr oduct)$
Here, $\Delta {h_f}$ is the enthalpy formation of any compound.
$\Delta {h_f} = (\dfrac{1}{2}\Delta {h_B}({H_2}) + \dfrac{1}{2}\Delta {h_B}(C{l_2})) - (\Delta {h_{B(HCl)}})$
here, $\Delta {h_B}({H_2})$ is the enthalpy formation of ${H_2}$,
$\Delta {h_B}(C{l_2})$ is the enthalpy formation of $C{l_2}$,
$\Delta {h_B}_{(HCl)}$ is the enthalpy formation of $HCl$ .
$\Rightarrow \Delta {h_f} = (\dfrac{1}{2}(104) + \dfrac{1}{2}(58)) - (103)$
$\Rightarrow \Delta {h_f} = 52 + 29 - 103$
$\therefore \Delta {h_f} = - 22\,kcal$
Hence, the correct option is B. -22.0 $kcal$ .
Note:Enthalpy of formation is estimated in units of energy per measure of substance, generally expressed in kilojoule per mole, yet additionally in kilocalorie per mole, joule per mole or kilocalorie per gram (any blend of these units adjusting to the energy per mass or sum rule).
Complete step-by-step solution:Bond dissociation energy is the energy required to break a chemical bond. It is one method for measuring the strength of a compound bond. Bond separation energy approaches bond energy just for diatomic atoms. The most grounded bond separation energy is for the Si-F bond. The most fragile energy is for a covalent bond and is similar.
The enthalpy of formation is the standard reaction enthalpy for the formation of the compound from its elements (atoms or molecules) in their most stable reference states at the chosen temperature (298.15K) and at 1bar pressure.
First, we’ll write the balanced reaction between the given compounds ${H_2}$ , $C{l_2}$ and $HCl$
$\dfrac{1}{2}{H_2} + \dfrac{1}{2}C{l_2} \to HCl$
According to the above reaction, we have the formula to find the enthalpy formation for the product $HCl$ gas:
$\Delta {h_f} = (Bond\,energy\,of\,reac\tan t) - (Bond\,energy\,of\,\Pr oduct)$
Here, $\Delta {h_f}$ is the enthalpy formation of any compound.
$\Delta {h_f} = (\dfrac{1}{2}\Delta {h_B}({H_2}) + \dfrac{1}{2}\Delta {h_B}(C{l_2})) - (\Delta {h_{B(HCl)}})$
here, $\Delta {h_B}({H_2})$ is the enthalpy formation of ${H_2}$,
$\Delta {h_B}(C{l_2})$ is the enthalpy formation of $C{l_2}$,
$\Delta {h_B}_{(HCl)}$ is the enthalpy formation of $HCl$ .
$\Rightarrow \Delta {h_f} = (\dfrac{1}{2}(104) + \dfrac{1}{2}(58)) - (103)$
$\Rightarrow \Delta {h_f} = 52 + 29 - 103$
$\therefore \Delta {h_f} = - 22\,kcal$
Hence, the correct option is B. -22.0 $kcal$ .
Note:Enthalpy of formation is estimated in units of energy per measure of substance, generally expressed in kilojoule per mole, yet additionally in kilocalorie per mole, joule per mole or kilocalorie per gram (any blend of these units adjusting to the energy per mass or sum rule).
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