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Question: \[\]\(\Delta H\) for the reaction, \(O{{F}_{2}}+{{H}_{2}}O\to {{O}_{2}}+2HF\): (B.E. of \[O-F\text{ ...

$$$$$\Delta Hforthereaction,for the reaction,O{{F}{2}}+{{H}{2}}O\to {{O}_{2}}+2HF$: (B.E. of O−F , O−H , H−FO-F\text{ },~O-H\text{ },~H-F and O=O O=O~ are 44, 111, 13544,\text{ }111,\text{ }135 and 119 kcal⋅mol−1119\text{ }kcal\cdot mo{{l}^{-1}} respectively)
A. −222 kcal~-222\text{ }kcal
B.−88 kcal-88\text{ }kcal
C.−111 kcal-111\text{ }kcal
D.−79 kcal-79\text{ }kcal

Explanation

Solution

We know that the identify for which bond of reactant will break as well as find their bond enthalpy which sums up to bond enthalpies of being in value for broken bond. By which we have to identify new bonds in the form of product and note down their negative bond enthalpy. Also note that we switch signs for bond enthalpies value to find energy released when bond form and this add up bond enthalpy value for formed product bond. Thus, by combining total value for breaking bond as well as forming bonds to acquire enthalpy of reaction.

Complete step-by-step answer: We have the reaction for which we have to find Enthalpy ΔH\Delta H for the reaction;
OF2+H2O→O2+2HFO{{F}_{2}}+{{H}_{2}}O\to {{O}_{2}}+2HF
Also binding energy for some bonds are given:
O−F bond = 44 kcalO-F\text{ }bond\text{ }=\text{ }44\text{ }kcal
O−H bond = 111 kcalO-H\text{ }bond\text{ }=\text{ }111\text{ }kcal
H−F bond = 135 kcalH-F\text{ }bond\text{ }=\text{ }135\text{ }kcal
O=O bond = 119 kcalO=O\text{ }bond\text{ }=\text{ }119\text{ }kcal
In order to find the enthalpies of reaction can be determine using five simple steps:
Step 1: We have to identify the broken bonds: We have that the OF2O{{F}_{2}} is already broken and since there is two O−FO-F bond in molecule by then we will add bond of enthalpies is as twice as. We also know that the H2O{{H}_{2}}O is too broken and it since, H2O{{H}_{2}}O has two H−OH-O bonds, we have to repeat the same process again and again.
Step 2: Now that broken bonds are being identified, now we have to find total energy to break bonds: Energy added to break bonds can be given as;
=[2×B.E.(O−F)]+[2×B.E.(O−H)]=[2\times B.E.(O-F)]+[2\times B.E.(O-H)]
By substituting the values of O−FO-F bond and O−HO-H bond we get;
=[(2×44)+(2×11)]=[(2\times 44)+(2\times 11)]
=88+22=88+22
⇒310kcal/mol\Rightarrow 310kcal/mol
Step 3: We have to identify formed bonds: From the reaction we know that the oxygen molecule is also formed that the one O+OO+O bond as well as two HF molecules are formed and have H−FH-F bonds in itself. Also since two molecule of HFHF is formed and also it will emit twice energy to form one molecule of HFHF
Step 4: Find total energy released to form new bond: Energy released to make product bonds,
=[2×B.E.(O=O)]+[2×B.E.(H−F)]=[2\times B.E.(O=O)]+[2\times B.E.(H-F)]
By substituting the values of O=OO=O bond and H−FH-F bond we get;
=[(2×44)+(2×11)]=[(2\times 44)+(2\times 11)]
=119+(2×135)=119+(2\times 135)
=119+270=119+270
⇒389kcal⋅mol−1\Rightarrow 389kcal\cdot mo{{l}^{-1}}
Now we know that this energy is released and the amount of released energy is given with negative sign.
⇒−389kcal⋅mol−1\Rightarrow -389kcal\cdot mo{{l}^{-1}}
Step 5: Final step is add up the energy for bond broken and formed:
ΔH= energy added to break reactant bonds + energy released when making product bonds\Delta H=\text{ }energy\text{ }added\text{ }to\text{ }break\text{ }reactant\text{ }bonds\text{ }+\text{ }energy\text{ }released\text{ }when\text{ }making\text{ }product\text{ }bonds
ΔH=310-389kcal⋅mol−1=−79kcal⋅mol−1\Delta H=\text{310-389kcal}\cdot \text{mo}{{\text{l}}^{-1}}=-79\text{kcal}\cdot \text{mo}{{\text{l}}^{-1}}
⇒−79kcal⋅mol−1\Rightarrow -79\text{kcal}\cdot \text{mo}{{\text{l}}^{-1}}

Hence, the correct option is D i.e. ΔH\Delta H is −79kcal-79kcal.

Note: Bond enthalpy as well as the enthalpy of the reaction will help us to understand how the chemical system uses the energy during reaction. Also the bond enthalpy describes how much energy needed for breaking or forming bonds, and it is also a measure of bond strength. Then by combining bond enthalpy value for all bonds broken and formed during reaction it's possible to estimate total change in potential energy of the system, which is ΔH\Delta H for reaction at constant pressure.