Question

2N2 + 5O2 = 2N2O5 at 25°C, the heat change at constant pressure is found to be -50160J. Calculate the value of internal energy change (∆u).
∆H = ∆uNgrt
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Answer 1

Answer:

Explanation:

Given equation

2N2 + 5O2= 2N2O5

we get,

• Delta H= -50160 J ( heat at constant pressure)
• delta n = number of moles of product - number of moles of reactants

                      = 2 - (5 + 2 ) =2-7 = -5

• T = 25+273 = 298 K
• R = 8.314 J
• delta U= ?

so now ,

we know that

delta H = delta U + delta n R T

-50160 = delta U + (-5) (8.314) ( 298)

delta U = - 37772.14 J