For the following reaction, what is the approximate value of Delta H?
CH4(g) --> CH3(g) + H(g)
I figured it out..
C-H = 411
4 moles on one side
3 on the other
so [4*411]-[3*411] = 411
To determine the approximate value of ΔH (enthalpy change) for the given reaction, CH4(g) → CH3(g) + H(g), you need to consider the enthalpies of the bonds broken and formed during the reaction.
In this case, you correctly identified that there are four C-H bonds broken on the reactant side (CH4) and three C-H bonds formed on the product side (CH3 and H). However, you made a calculation error in determining the value of ΔH.
The enthalpy change for a reaction can be calculated by subtracting the total energy required to break the bonds from the total energy released when new bonds are formed. In other words, ΔH = (Energy absorbed - Energy released).
To correct the calculation, you need to multiply the number of bonds by the bond energy value. The given C-H bond energy value of 411 kJ/mol is the average bond energy for a C-H bond.
On the reactant side (CH4), there are four C-H bonds, so the energy required to break these bonds is 4 * 411 kJ/mol.
On the product side, there are three C-H bonds (from CH3) and one H-H bond. To simplify the calculation, let's assume the bond energy of C-H is the same in CH3 as it is in CH4. Thus, the energy released when the bonds are formed is (3 * 411 kJ/mol) + (1 * H-H bond energy).
The bond energy of H-H (hydrogen-hydrogen) is typically around 432 kJ/mol.
To calculate ΔH, subtract the energy required to break the bonds from the energy released when the bonds are formed:
ΔH = (4 * 411 kJ/mol) - ((3 * 411 kJ/mol) + (1 * 432 kJ/mol)).
After performing the calculation, you will get the approximate value of ΔH for the given reaction.