Question

Relation betn heat of reaction at constant temp & pressure

Answer 1

HELLO THERE!

First, let us check Heat of reaction at Constant temperature:

From the First Law of Thermodynamics, we know:

du = dq + dw

Where du = Internal energy, dq = Heat, dw = Work!

When temperature is constant, du = 0, as:

($du = nC_{v} \triangle T$

Since temperature is constant, ΔT = 0, so du = 0.)

So, we get dq = - dw

Now, dw = - PdV

So, dq = PdV.

Now, let us check Heat of Reaction at Constant pressure:

We know, that:

$\triangle H = du + PdV + VdP\\\\or, \triangle H = (dq - PdV) + PdV + VdP\\\\or, \triangle H = dq + VdP$

Since pressure is constant, dP = 0, so VdP = 0, and we get:

dq = dH (where dH = change in enthalpy).

These were the relations:

At constant temperature,

dq = PdV

At constant pressure,

dq = dH

At constant volume,

dq = dU (not asked in your question, so I'm not explaining how).

THANKS!