Calculating ∆rH° for Cu + O2 -> CuO

[Diamond101]

Homework Statement

1. (a) From the following data, 2 Cu (s) + S (s) Cu2S (g) ∆rH° = - 79.5 kJ mol-1 S (s) + O2(g) SO2 (g) ∆rH° = - 296.8 kJ mol-1 Cu2S (s) + 2 O2(g) 2CuO (s) + SO2(g) ∆rH° = - 527.5 kJ mol-1 Determine ∆rH° for Cu(s) + ½ O2 (g) CuO (s) at 298 K

(b) Given the following Cp / (J K-1 mol-1) Cu(s) = 24.44 O2(g) = 29.36 CuO (s) = 42.30 Calculate the ∆rH° for the reaction at 1000K.

Homework Equations

More so part b is it correct

The Attempt at a Solution

2 Cu (s) + S (s) ---> Cu2S (g) ∆rH° = - 79.5 kJ mol-1
SO2 (g) ---> S (s) + O2(g) ∆rH° = +296.8 kJ mol-1
Cu2S (s) + 2 O2(g) ---> 2CuO (s) + SO2(g) ∆rH° = - 527.5 kJ mol-1

Adding the above three equations results in this:

2 Cu (s) + O2(g) ---> 2CuO (s) ∆rH° = -310.2 kJ

Dividing through by 2 = -155.1 kJ.
part b cu(s) =+79.5 =24.4 (1000-298 k) using the equation ∆rHT2=∆rHT1+∆CP(T2-T1)
SAME GOES FOR O2 and CUO ?

 

[CrazyNinja]

How did you get the last equation? ΔH(T1)=ΔH(T2) + C_pΔT ?

 

[Diamond101]

actually i got Δcp= cp products -reactants= 43.3-24.44+0.5*29.36)= 3.18 kj/mol
ΔHr 1000= ΔHr 298+ΔCpΔT
-155.1*10^3 J-MOL * 3.18 JK-MOL *702 K
=-155.1 *10^3 + 2232.36
= 152.9 KJ/MOL

 

[Diamond101]

How did you get the last equation? ΔH(T1)=ΔH(T2) + C_pΔT ?

And the equation was my lecture notes

 

[Diamond101]

How did you get the last equation? ΔH(T1)=ΔH(T2) + C_pΔT ?

actually i got Δcp= cp products -reactants= 43.3-24.44+0.5*29.36)= 3.18 kj/mol
ΔHr 1000= ΔHr 298+ΔCpΔT
-155.1*10^3 J-MOL * 3.18 JK-MOL *702 K
=-155.1 *10^3 + 2232.36
= 152.9 KJ/MOL

 

[CrazyNinja]

Check the equation once again. It is dimensionally incorrect.

C_p has units J K^-1 mol^-1 while ΔH has units J mole^-1