estimate the enthalpy change for the following reaction:
CH3Cl(g) + Cl2(g) CH2Cl2(g) + HCl(g)
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So you have to add CH3Cl(g) with Cl2(g) then minus (CH2CL2(g) + HCl(G)) in order to find the enthalpy change...
I look at the answer and it says that the bond energy is 413 kj per mole for CH3Cl
Note that 413 kj is the energy for the bond of C-H, however, there are 3 bonds of this type (for CH3CL), and there is also the C-Cl bond.
So why is it 413 instead of 3(413)+BE(C-Cl)?
CH3Cl(g) + Cl2(g) CH2Cl2(g) + HCl(g)
-------
So you have to add CH3Cl(g) with Cl2(g) then minus (CH2CL2(g) + HCl(G)) in order to find the enthalpy change...
I look at the answer and it says that the bond energy is 413 kj per mole for CH3Cl
Note that 413 kj is the energy for the bond of C-H, however, there are 3 bonds of this type (for CH3CL), and there is also the C-Cl bond.
So why is it 413 instead of 3(413)+BE(C-Cl)?
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You're breaking a single C-H bond and a single Cl-Cl bond, and forming a single C-Cl bond and a single H-Cl bond. So the enthalpy change will be ΔH(C-Cl) + ΔH(H-Cl) - ΔH(C-H) - ΔH(Cl-Cl)
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fantasy