Given the enthalpy changes for the following reactions: 2W(s) + 3O2(g) ---> 2WO(

Question

Given the enthalpy changes for the following reactions:

2W(s) + 3O2(g) ---> 2WO(s) DeltaH= -1680kj

C(s) + O2(g) --> CO2(g) DeltaH = -393.5 kj

2WC(s) _ 5O2(g) ---. 2WO3(s) + 2CO2(g) DeltaH = -2391.6 kj

determine the enthalpy change for the following reaction:

W(s) + C9s) ---> WC(s)

a) +317.5kj

b) -38.0 kj

c) -317.5 kj

d) -1488.0 kj

e) +1488.0 kj

The answer is B but I haven't the lishgtest idea the steps on how to get to that answer. If you would please, show your steps I wold be VERY appreciative! Thank you!

Explanation / Answer

2W(s) + 3O2(g) ---> 2WO(s) DeltaH1= -1680kj ----------(1)

C(s) + O2(g) --> CO2(g) DeltaH2 = -393.5 kj ---------------(2)

2WC(s) _ 5O2(g) ---. 2WO3(s) + 2CO2(g) DeltaH3 = -2391.6 kj -----------(3)

[1/2*Eq(1)] + Eq(2) - [1/2*Eq(3)]

we get,

W(s) + C(s) ---> WC(s)    delta H = ?

delta H =  [1/2*DeltaH1] +  [DeltaH2] - [1/2* DeltaH3]

= [1/2* (-1680kj)] + [ -393.5 kj] - [ 1/2*(-2391.6 kj)]

= -37.7kj   

Almost b) -38.0 kj

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