`=>` The heat absorbed at constant volume is equal to change in the internal energy i.e., `color{purple}(ΔU = q_V)`.
● But most of chemical reactions are carried out not at constant volume, but in flasks or test tubes under constant atmospheric pressure. So, we need to define another state function which may be suitable under these conditions.
`=>` We may write equation (6.1) as `color{purple}(ΔU = q_p− pΔV)` at constant pressure, where `q_p` is heat absorbed by the system and `color{purple}(–pΔV)` represent expansion work done by the system.
● Let us represent the initial state by subscript `1` and final state by `2`. We can rewrite the above equation as
`color{purple}(U_2–U_1 = q_p – p (V_2 – V_1)`
● On rearranging, we get
`color{purple}(q_p = (U_2 + pV_2) – (U_1 + pV_1))` ........(6.6)
Now we can define another thermodynamic function, the enthalpy `H` [Greek word enthalpien, to warm or heat content] as :
`color{purple}(H = U + pV)` .................... (6.7)
So, equation (6.6) becomes
`color{purple}(q_p= H_2 – H_1 = ΔH)`
`=>` Although `color{purple}(q)` is a path dependent function, `color{purple}(H)` is a state function because it depends on `U`, `p` and `V`, all of which are state functions.
● Therefore, `color{purple}(ΔH)` is independent of path. Hence, `color{purple}(q_p)` is also independent of path.
`=>` For finite changes at constant pressure, we can write equation 6.7 as
`color{purple}(ΔH = ΔU + ΔpV)`
Since `color{purple}(p)` is constant, we can write
`color{purple}(ΔH = ΔU + pΔV)` .................... (6.8)
`color{red}("Note ")` When heat is absorbed by the system at constant pressure, we are actually measuring changes in the enthalpy.
`=>` `color{purple}(ΔH = q_p)`, heat absorbed by the system at constant pressure.
● `color{purple}(ΔH)` is negative for `color{red}("exothermic reactions")` which evolve heat during the reaction
● `color{purple}(ΔH)` is positive for `color{red}("endothermic reactions")` which absorb heat from the surroundings.
`=>` At constant volume `color{purple}((ΔV = 0), ΔU = q_V),` therefore equation 6.8 becomes
`color{purple}(ΔH = ΔU = q_V)`
`=>` The difference between `color{purple}(ΔH)` and `color{purple}(ΔU)` is not usually significant for systems consisting of only solids and/or liquids. Solids and liquids do not suffer any significant volume changes upon heating.
`=>` The difference becomes significant when gases are involved.
● Let us consider a reaction involving gases.
● If `color{purple}(V_A)` is the total volume of the gaseous reactants, `color{purple}(V_B)` is the total volume of the gaseous products, `n_A` is the number of moles of gaseous reactants and `color{purple}(n_B)` is the number of moles of gaseous products, all at constant pressure and temperature, then using the ideal gas law, we write,
`color{purple}(pV_A = n_ART)` and `color{purple}(pV_B = n_BRT)`
Thus, `color{purple}(pV_B – pV_A = n_BRT – n_ART = (n_B–n_A)RT)`
or `color{purple}(p (V_B – V_A) = (n_B – n_A) RT)`
or` color{purple}(p ΔV = Δn_gRT)` ................. (6.9)
● Here, `color{purple}(Δn_g)` refers to the number of moles of gaseous products minus the number of moles of gaseous reactants.
● Substituting the value of `color{purple}(pΔV)` from equation 6.9 in equation 6.8, we get
`color{purple}(ΔH = ΔU + Δn_gRT)` ................ (6.10)
The equation 6.10 is useful for calculating `color{purple}(ΔH)` from `color{purple}(ΔU)` and vice versa.
`=>` The heat absorbed at constant volume is equal to change in the internal energy i.e., `color{purple}(ΔU = q_V)`.
● But most of chemical reactions are carried out not at constant volume, but in flasks or test tubes under constant atmospheric pressure. So, we need to define another state function which may be suitable under these conditions.
`=>` We may write equation (6.1) as `color{purple}(ΔU = q_p− pΔV)` at constant pressure, where `q_p` is heat absorbed by the system and `color{purple}(–pΔV)` represent expansion work done by the system.
● Let us represent the initial state by subscript `1` and final state by `2`. We can rewrite the above equation as
`color{purple}(U_2–U_1 = q_p – p (V_2 – V_1)`
● On rearranging, we get
`color{purple}(q_p = (U_2 + pV_2) – (U_1 + pV_1))` ........(6.6)
Now we can define another thermodynamic function, the enthalpy `H` [Greek word enthalpien, to warm or heat content] as :
`color{purple}(H = U + pV)` .................... (6.7)
So, equation (6.6) becomes
`color{purple}(q_p= H_2 – H_1 = ΔH)`
`=>` Although `color{purple}(q)` is a path dependent function, `color{purple}(H)` is a state function because it depends on `U`, `p` and `V`, all of which are state functions.
● Therefore, `color{purple}(ΔH)` is independent of path. Hence, `color{purple}(q_p)` is also independent of path.
`=>` For finite changes at constant pressure, we can write equation 6.7 as
`color{purple}(ΔH = ΔU + ΔpV)`
Since `color{purple}(p)` is constant, we can write
`color{purple}(ΔH = ΔU + pΔV)` .................... (6.8)
`color{red}("Note ")` When heat is absorbed by the system at constant pressure, we are actually measuring changes in the enthalpy.
`=>` `color{purple}(ΔH = q_p)`, heat absorbed by the system at constant pressure.
● `color{purple}(ΔH)` is negative for `color{red}("exothermic reactions")` which evolve heat during the reaction
● `color{purple}(ΔH)` is positive for `color{red}("endothermic reactions")` which absorb heat from the surroundings.
`=>` At constant volume `color{purple}((ΔV = 0), ΔU = q_V),` therefore equation 6.8 becomes
`color{purple}(ΔH = ΔU = q_V)`
`=>` The difference between `color{purple}(ΔH)` and `color{purple}(ΔU)` is not usually significant for systems consisting of only solids and/or liquids. Solids and liquids do not suffer any significant volume changes upon heating.
`=>` The difference becomes significant when gases are involved.
● Let us consider a reaction involving gases.
● If `color{purple}(V_A)` is the total volume of the gaseous reactants, `color{purple}(V_B)` is the total volume of the gaseous products, `n_A` is the number of moles of gaseous reactants and `color{purple}(n_B)` is the number of moles of gaseous products, all at constant pressure and temperature, then using the ideal gas law, we write,
`color{purple}(pV_A = n_ART)` and `color{purple}(pV_B = n_BRT)`
Thus, `color{purple}(pV_B – pV_A = n_BRT – n_ART = (n_B–n_A)RT)`
or `color{purple}(p (V_B – V_A) = (n_B – n_A) RT)`
or` color{purple}(p ΔV = Δn_gRT)` ................. (6.9)
● Here, `color{purple}(Δn_g)` refers to the number of moles of gaseous products minus the number of moles of gaseous reactants.
● Substituting the value of `color{purple}(pΔV)` from equation 6.9 in equation 6.8, we get
`color{purple}(ΔH = ΔU + Δn_gRT)` ................ (6.10)
The equation 6.10 is useful for calculating `color{purple}(ΔH)` from `color{purple}(ΔU)` and vice versa.