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Question: The reaction \[C{H_{4(g)}} + C{l_{2(g)}} \to C{H_3}C{l_{(g)}} + HC{l_{(g)}}\] has \[\Delta H = - 25k...

The reaction CH4(g)+Cl2(g)CH3Cl(g)+HCl(g)C{H_{4(g)}} + C{l_{2(g)}} \to C{H_3}C{l_{(g)}} + HC{l_{(g)}} has ΔH=25kcal\Delta H = - 25kcal

BondBond Energy Kcal
CClC - Cl8484
HClH - Cl103103
CHC - Hx
ClClCl - Cly
x:y=9:5x:y = 9:5

From the given data what is bond energy of ClClCl - Cl bond
(1)$$$$70kcal
(2)$$$$80kcal
(3)$$$$67.75kcal
(4)$$$$57.75kcal

Explanation

Solution

Bond energy can be defined as the amount of energy that is necessary to form the bond or break the bond between two atoms. Energy is released by the system when a bond is formed and energy is given to the system when a bond is broken.

Complete answer:
Let’s write the given chemical question
CH4(g)+Cl2(g)CH3Cl(g)+HCl(g)C{H_{4(g)}} + C{l_{2(g)}} \to C{H_3}C{l_{(g)}} + HC{l_{(g)}}
From the above reaction we can see that a CHC - H bond is broken from the molecule of methane and one ClClCl - Cl bond is broken from the chlorine molecule. One the product side we can see the formation of two new bonds, one is CClC - Cl bond to form chloroform and other is HClH - Cl bond to form hydrochloric acid.
Let us write the given value of bond enthalpies of different bonds.
Bond enthalpy of Carbon-Chlorine bond, CCl=84kcalC - Cl = 84kcal
Bond enthalpy of Hydrogen-Chlorine bond, HCl=103kcalH - Cl = 103kcal
Bond enthalpy of Carbon-Hydrogen bond, CH=xkcalC - H = xkcal
Bond enthalpy of Chlorine-Chlorine bond, ClCl=ykcalCl - Cl = ykcal
Total energy released/given, ΔH=25kcal\Delta H = - 25kcal
Total released/given energy is given by measuring the difference between energy given during breaking bonds and energy released due to formation of bonds. Now using the formula for total released/given energy according to the given reaction.
ΔH=4×CH+1×ClCl3×CH1×CCl1×HCl\Delta H = 4 \times C - H + 1 \times Cl - Cl - 3 \times C - H - 1 \times C - Cl - 1 \times H - Cl
Substituting given values in the above equation.
25=4x+y3x84103- 25 = 4x + y - 3x - 84 - 103
On further simplification we get the relation
x+y=162x + y = 162
also, we are given x:y=9:5x:y = 9:5 which can be written as x=95yx = \dfrac{9}{5}y
putting this value of x in the obtained relation
95y+y=162\dfrac{9}{5}y + y = 162
y=162×514y = \dfrac{{162 \times 5}}{{14}}
we get the value of y
y=57.75kcaly = 57.75kcal
So, the bond energy of ClClCl - Cl bond is 57.75kcal57.75kcal.

Therefore option (4)(4) is the right answer.

Note:
Laws of thermodynamics, states that bond enthalpy is positive when a bond is formed and bond enthalpy is negative when bond is broken. The negative sign of total energy released indicates that the energy released by the formation of new bonds is less than the energy required to break the bonds in reactant molecules.