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 for the C-H bond (412 kJ mol^-1), we need to consider the bonds broken and formed in each of the given processes.
A bond is broken when a compound is converted into its constituent elements or other compounds. A bond is formed when elements or compounds combine to create a new compound.
Let's analyze each option step-by-step:
A. CH₄(g) → C(s) + 2H₂(g)
In this process, one C-H bond is broken in CH₄, forming one C-C bond in C(s) and two H-H bonds in H₂. So, a total of 412 kJ is released from breaking the C-H bond, and 2(436 kJ) = 872 kJ is required to form the new H-H bonds. The net enthalpy change for process A is -872 + 412 = -460 kJ.
B. CH₄(g) → C(g) + 2H₂(g)
In this process, one C-H bond is broken, forming one C-C bond and two H-H bonds. Similar to process A, 412 kJ is released from breaking the C-H bond, and 2(436 kJ) = 872 kJ is required to form the new H-H bonds. The net enthalpy change for process B is -872 + 412 = -460 kJ.
C. CH₄(g) → C(s) + 4H(g)
In this process, four H-H bonds are broken, and only one C-H bond is broken. Four H-H bonds require 4(436 kJ) = 1744 kJ to break, and one C-H bond releases 412 kJ. The net enthalpy change for process C is -1744 + 412 = -1332 kJ.
D. CH₄(g) → CH₃(g) + H(g)
In this process, only one C-H bond is broken, and one C-H bond is formed. So, 412 kJ is released from breaking the C-H bond, and 412 kJ is required to form a new C-H bond. The net enthalpy change for process D is 0 kJ.
Comparing the net enthalpy changes for each process:
A. -460 kJ
B. -460 kJ
C. -1332 kJ
D. 0 kJ
Process A and B have the closest net enthalpy change (-460 kJ) to the average bond enthalpy for the C-H bond (412 kJ mol^-1). Therefore, the answer is either A or B, depending on how closely you define "closest."
To determine which process has an enthalpy change closest to the average bond enthalpy for the C-H bond, we need to calculate the change in enthalpy for each process and compare it to 412 kJ mol^-1.
First, let's analyze each process:
A. CH4(g) ---> C(s) + 2H2(g)
In this process, one C-H bond is broken and one C-C bond is formed. There are also two H-H bonds formed. The net change in bond enthalpy can be calculated as follows:
CH4(g): -1 C-H bond * 412 kJ mol^-1 = -412 kJ mol^-1
C(s): +1 C-C bond * average bond enthalpy of C-C bond
2H2(g): +2 H-H bonds * average bond enthalpy of H-H bond
B. CH4(g) ---> C(g) + 2H2(g)
In this process, one C-H bond is broken and one C-C bond is formed. There are also two H-H bonds formed. The net change in bond enthalpy can be calculated similar to process A, but note that the product C(g) does not have any bonds formed.
C. CH4(g) ---> C(s) + 4H(g)
In this process, one C-H bond is broken and one C-C bond is formed. There are also four H-H bonds formed. Similar to process A, calculate the net change in bond enthalpy.
D. CH4(g) ---> CH3(g) + H(g)
In this process, one C-H bond is broken and one C-C bond is formed. There is also one H-H bond formed.
Once we have calculated the net change in bond enthalpy for each process, we can compare it to the average bond enthalpy for the C-H bond, which is 412 kJ mol^-1. The process with the enthalpy change closest to 412 kJ mol^-1 is the answer.
(Note: To calculate the change in bond enthalpy, you would need to know the average bond enthalpies of the C-C and H-H bonds. These values are not provided in the question. So, to give you the precise answer, we need this information. However, you can follow the same approach and compare the net change in bond enthalpies to 412 kJ mol^-1 for each process given the relevant bond enthalpy values.)