Question 3 of 4 Incorrect IncorrectIncorrectIncorrectIncorrectIncorrect Incorrec

Question

Question 3 of 4 Incorrect IncorrectIncorrectIncorrectIncorrectIncorrect IncorrectIncorrect University Science BoOKS presented by Sapling Leaning Map do The oxidation of copper(l) oxide, Cu20(s), to copper(II) oxide, CuO(s), is an exothermic process. 2Cmo(s) +02 (g) -> 4CnO(s) The change in enthalpy upon reaction of 11.69 g of Cu20(s) is-11.93 kJ. Calculate the work, w, and energy change, AUrn,when 11.69 g of Cu2O(s) is oxidized at a constant pressure of 1.00 bar and a constant temperature of 25 °C Number Number Note that Erxn is sometimes used w=11-0.099 | kJ = 11-103.47 | kJl as the symbol for energy change instead of xn.

Explanation / Answer

DH = DU+PDV

   PDV = DnRT

So that,

DH = DU + DnRT

from reaction , Dn = 0-1 = -1

DnRT = -1*8.314*298 = -2.5 kj

pDV = -2.5 Kj

Work done(W)= - PDV

            = -(-2.5)

   = 2.5 kj
dU = DH-DnRT

   = -11.93-(-2.5)

DU = -9.43 kj

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