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How to find change in enthalpy in  such
typ of que1.i f the the pressure is constant change in enthalpy should eqeal heat transfer q as p∆v will get strike out. In constant volume p∆v is already going to be 0and enthalpy should eqeal q+v∆p but sir iIam not getting answer please correct my concect
9 months ago

HELLO THERE!

There are some important points to consider for answering this question.

dH = U + PdV
We also know that:
dH = TdS + VdP
Again, Heat (q) = TdS

So, dH = q + VdP
Hence for constant pressure, (dP = 0), heat of a reaction is equal to the change in enthalpy (dH).

One more thing to note:
From First Law of Thermodynamics, we have:
dq = du – dw = du – PdV
For constant volume, dV = 0, so q =dU.

In the above question, we have to calculate the heat of the reaction at constant Volume, which is equal to the Change in Internal Energy.
dH = dU + d(PV)
or, dU = dH – nRT

Change in moles = Moles of product – Moles of reactant = 6 – 7.5 = -1.5

R = 1.98 cal/mol K = 2 cal/mol K
So, dU = -780980 + (1.5 x 2 x 298) = 894 – 780980 cal = -780086 cal

Thanks!

9 months ago
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