CH4(g)+Cl2(g)---->CH3Cl(g)+HCl(g) A) Use bond energies to cal. delta H My Ans: - 110 KJ/mol B) Use...

CH4(g)+Cl2(g)---->CH3Cl(g)+HCl(g)

A) Use bond energies to cal. delta H
My Ans: - 110 KJ/mol
B) Use the enthalpy of formation to cal. delta H
My Ans: 103 KJ/mol
C) Explain why A and B are not the same?
D) Calculate % error?

Expert Solution

Since you have calculated answers of part A and part B. So I am solving part C and part D here:

C) A and B are not same because while calculating bond energies we take the average value.

Like if we talk about a C-H bond then average value of all the C-H bonds in all the known molecule will be the values of bond energy of C-H bond. Now if we use this value to calculate H of reaction then it will surely differ form the value calculated using enthalpy of formation because enthalpy of formation of every molecule is different.

D) % error = (Difference in values / True value) x 100

= (7/103) x 100

= 6.8%

queen_honey_blossom answered 2 years ago

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