ΔH0 for CO at 60°C is given as – 285200 kJ/mole. Calculate ΔH0 at 2500°C given that the enthalpies of gases concerned in kJ/mole are as follows :

Question

[latex]\Delta H_0[/latex] for CO at 60°C is given as – 285200 kJ/mole. Calculate [latex]\Delta H_0[/latex] at 2500°C given that the enthalpies of gases concerned in kJ/mole are as follows :

Gas	60°C	2500°C
CO	9705	94080
O	9696	99790
[latex]CO _2[/latex]	10760	149100

Accepted Answer

The reaction equation is given by

[latex]CO +\frac{1}{2} O _2 \longrightarrow CO _2[/latex]

Refer Fig. 11.7.

It can be seen from the property diagram of Fig. 11.7 that the enthalpy of combustion at temperature [latex]T, \Delta H_T[/latex] can be obtained from [latex]\Delta H_0 ext { and } T_0[/latex] by the relationship

[latex]-\Delta H_T=-\Delta H_0+\left(H_{R_T}-H_{R_0} ight)-\left(H_{P_T}-H_{P_0} ight)[/latex] ...(i)

where [latex]H_{R_T}-H_{R_0}=[/latex] increase in enthalpy of the reactants from [latex]T_0 ext { to } T[/latex]

and [latex]H_{P_T}-H_{P_0}=[/latex] increase in enthalpy of the products from [latex]T_0 ext { to } T[/latex]

Now, from the given data, we have

[latex]H_{R_0}=1 imes 9705+\frac{1}{2} imes 9696=14553[/latex] kJ

[latex]H_{R_T}=1 imes 94080+\frac{1}{2} imes 99790=143975[/latex], kJ

[latex]H_{P_0}=1 imes 10760=10760[/latex] kJ

[latex]H_{P_T}=1 imes 149100[/latex] kJ = 149100 kJ

Using equation (i), we get

[latex]-\Delta H_T=+285200+(143975-14553)-(149100-10760)[/latex]

= 285200 + 129422 – 138340 = 276282

∴ [latex]\Delta H _{ T }=-276282[/latex] kJ/mole.

Question 11.33: ΔH0 for CO at 60°C is given as – 285200 kJ/mole. Calculate Δ...

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