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Question: The bond dissociation energy of gaseous \({H_2}\) , \(C{l_2}\) and \(HCl\) are 104, 58 and 103 \(kca...

The bond dissociation energy of gaseous H2{H_2} , Cl2C{l_2} and HClHCl are 104, 58 and 103 kcalmol1kcal\,mo{l^{ - 1}} respectively. The enthalpy formation for HClHCl gas will be:
A. -44.0 kcalkcal
B. -22.0 kcalkcal
C. 22.0 kcalkcal
D. 44.0 kcalkcal

Explanation

Solution

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 HClHCl 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 H2{H_2} , Cl2C{l_2} and HClHCl
12H2+12Cl2HCl\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 HClHCl gas:
Δhf=(Bondenergyofreactant)(BondenergyofProduct)\Delta {h_f} = (Bond\,energy\,of\,reac\tan t) - (Bond\,energy\,of\,\Pr oduct)
Here, Δhf\Delta {h_f} is the enthalpy formation of any compound.
Δhf=(12ΔhB(H2)+12ΔhB(Cl2))(ΔhB(HCl))\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, ΔhB(H2)\Delta {h_B}({H_2}) is the enthalpy formation of H2{H_2},
ΔhB(Cl2)\Delta {h_B}(C{l_2}) is the enthalpy formation of Cl2C{l_2},
ΔhB(HCl)\Delta {h_B}_{(HCl)} is the enthalpy formation of HClHCl .
Δhf=(12(104)+12(58))(103)\Rightarrow \Delta {h_f} = (\dfrac{1}{2}(104) + \dfrac{1}{2}(58)) - (103)
Δhf=52+29103\Rightarrow \Delta {h_f} = 52 + 29 - 103
Δhf=22kcal\therefore \Delta {h_f} = - 22\,kcal

Hence, the correct option is B. -22.0 kcalkcal .

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).