show: ΔH = qp

enthalpy, H, is defined as:

H = E + PV

"taking the derivative"  [actually, should evalute the "differential"]

dH = dE + d(PV)

 = dE + PdV + VdP [1].

For a constant pressure process, dP = 0; thus [1] becomes

 dH = dE + PdV [2].

Work, W, is defined as:

W = - F * x

 or dW = - F dx [3]

where F = force and x= displacement.

Pressure, P, is defined as:

P = F / A

 or F = P A [4]

substituting [4] into [3],

 dW = - P A dx = - P dV [5]

substituting [5] into [2],

 dH = dE - dW [6].

The 1st law of thermodynamics,

 dE = dq + dW [7]

substituting [7] into [6],

dH = dq + dW - dW

dH = dq

hence,

ΔH = qp.

show ΔG < 0 for a spontaneous process.

The 2nd law of thermodynamics states that for a spontaneous process,

 ΔSuniverse = Δ Ssystem + Δ Ssurrounding > 0 [1].

As

qsystem = - qsurrounding

and at constant pressure (see above),

qsystem = Δ H

thus

qsurrounding = - Δ H.

Dividing by T and using the definition of Δ S,

 Δ Ssurrounding = qsurrounding / T = - Δ H / T [2]

substituting [2] into [1],

Δ Suniverse = Δ Ssystem + - Δ H / T > 0

multiplying by - T,

- T Δ Suniverse = - T Δ Ssystem + Δ H < 0.

Defining the change in Gibbs Free Energy as:

Δ G = - T Δ Suniverse

substituting and rearranging,

Δ G = Δ H - T Δ S < 0 for a spontaneous process.