use the following data to calculate the mean bond enthalpy values for the C-H and C-C bonds.
CH4(g)-C(g) + 4H(g) standard enthalpy change is +1648 kj/mol
C2H6(g)-2C(g) + 6H(g) standard enthalpy change is +2820 kj/mol
412 kj mol^-1 (4 bonds and they're all C-H so it's 1648 /4)
and 2820-(6*412) = 348 kj mol^-1
remember there aren't 2 C-C bonds - just the one!!!
To calculate the mean bond enthalpy values for the C-H and C-C bonds, we need to use the concept of Hess's Law. Using the given data, we can write the following equations:
1) CH4(g) → C(g) + 4H(g)
2) C2H6(g) → 2C(g) + 6H(g)
By reversing equation 1), we have:
3) C(g) + 4H(g) → CH4(g)
We can subtract equation 3) from equation 2) to get the desired equation:
4) C2H6(g) - CH4(g) → C(g)
Now, let's analyze the enthalpy changes:
Equation 2) states that the standard enthalpy change is +2820 kJ/mol.
Equation 1) states that the standard enthalpy change is +1648 kJ/mol.
Since equation 4) is the difference between equation 2) and equation 3), we subtract the enthalpy changes:
ΔH4 = ΔH2 - ΔH1
ΔH4 = (+2820 kJ/mol) - (+1648 kJ/mol)
ΔH4 = +1172 kJ/mol
Now, we can calculate the mean bond enthalpy values:
Mean C-H bond enthalpy = ΔH4 / Number of H atoms
Mean C-H bond enthalpy = (+1172 kJ/mol) / 4 H atoms
Mean C-H bond enthalpy = +293 kJ/mol
Mean C-C bond enthalpy = ΔH4 / Number of C atoms
Mean C-C bond enthalpy = (+1172 kJ/mol) / 1 C atom
Mean C-C bond enthalpy = +1172 kJ/mol
Therefore, the mean bond enthalpy values for the C-H and C-C bonds are approximately +293 kJ/mol and +1172 kJ/mol, respectively.
To calculate the mean bond enthalpy values for the C-H and C-C bonds using the given data, you need to use the concept of the enthalpy change of a reaction.
The enthalpy change of a reaction can be expressed as the sum of the bonds broken minus the sum of the bonds formed. In this case, we have two reactions and we want to determine the bond enthalpies of the C-H and C-C bonds.
Let's consider the first reaction:
CH4(g) - C(g) + 4H(g) (Enthalpy change: +1648 kJ/mol)
In this reaction, one C-H bond and one C-C bond are broken. The bond enthalpy of the C-H bond is represented by ΔH(CH) and the bond enthalpy of the C-C bond is represented by ΔH(CC).
Therefore, we can write the enthalpy change equation for this reaction:
ΔH(CH4) = ΔH(CH) + ΔH(CC) + 4ΔH(H)
Substituting the given values:
1648 kJ/mol = ΔH(CH) + ΔH(CC) + 4ΔH(H) -------- (Equation 1)
Next, let's consider the second reaction:
C2H6(g) - 2C(g) + 6H(g) (Enthalpy change: +2820 kJ/mol)
In this reaction, two C-H bonds and one C-C bond are broken. So the enthalpy change equation for this reaction is:
ΔH(C2H6) = 2ΔH(CH) + ΔH(CC) + 6ΔH(H)
Substituting the given values:
2820 kJ/mol = 2ΔH(CH) + ΔH(CC) + 6ΔH(H) -------- (Equation 2)
Now, we have a system of two equations (Equation 1 and Equation 2) with two unknowns (ΔH(CH) and ΔH(CC)). We can solve this system of equations to find the values of ΔH(CH) and ΔH(CC), which represent the mean bond enthalpies for the C-H and C-C bonds, respectively.
By solving the system of equations, you can obtain the numerical values for ΔH(CH) and ΔH(CC), which will give you the mean bond enthalpy values for the C-H and C-C bonds, respectively.