Please explain.

The average bond enthalpy for the C-H bond is 412 kJ mol^-1. Which process has an enthalpy change closest to this value ?

A. CH 4 (g)---> C(s)+2H2(g)
B. CH 4(g)--> C(g)+2H2(g)
C. CH 4(g)---> C(s)+ 4H(g)
D. CH 4 (g)---> CH3(g)+ H(g)

To determine which process has an enthalpy change closest to the average bond enthalpy of the C-H bond (412 kJ mol^-1), we need to identify the total energy change required to break and form the bonds involved in each process.

Let's analyze each option:

A. CH 4(g) ---> C(s) + 2H2(g)

In this process, we are breaking four C-H bonds and forming one C-C bond and four H-H bonds. The total energy change can be calculated as follows:

Total energy change = (4 x bond enthalpy of C-H) - (bond enthalpy of C-C) - (4 x bond enthalpy of H-H)

Total energy change = (4 x 412 kJ mol^-1) - 0 - (4 x 436 kJ mol^-1)

B. CH 4(g) ---> C(g) + 2H2(g)

In this process, we are breaking four C-H bonds and forming one C-C bond and four H-H bonds. The total energy change is:

Total energy change = (4 x bond enthalpy of C-H) - (bond enthalpy of C-C) - (4 x bond enthalpy of H-H)

Total energy change = (4 x 412 kJ mol^-1) - 0 - (4 x 436 kJ mol^-1)

C. CH 4(g) ---> C(s) + 4H(g)

In this process, we are breaking four C-H bonds and forming one C-C bond and four H-H bonds. The total energy change is:

Total energy change = (4 x bond enthalpy of C-H) - (bond enthalpy of C-C) - (4 x bond enthalpy of H-H)

Total energy change = (4 x 412 kJ mol^-1) - 0 - (4 x 436 kJ mol^-1)

D. CH 4(g) ---> CH3(g) + H(g)

In this process, we are breaking four C-H bonds and forming three C-H bonds and one H-H bond. The total energy change is:

Total energy change = (4 x bond enthalpy of C-H) - (3 x bond enthalpy of C-H) - (bond enthalpy of H-H)

Total energy change = (4 x 412 kJ mol^-1) - (3 x 412 kJ mol^-1) - (436 kJ mol^-1)

After calculating the total energy changes for each process, we can compare them to the average bond enthalpy of the C-H bond (412 kJ mol^-1). The process with the total energy change closest to 412 kJ mol^-1 will be the answer.

Therefore, by comparing the calculations, we find that the process closest to the average bond enthalpy of the C-H bond is option C) CH4(g) ---> C(s) + 4H(g).