Question

Given : Rg2O (s) + 1/2 O2 (g) -----> 2RgO (s) ΔH°= -141 kJ
and Rg2O (s) ------> Rg (s) + RgO (s) ΔH° = +11 kJ
The standard enthalpy of formation of RgO (s) if nearest
a. -163 kJ
b. -300 kJ
c. +300 kJ
d. +150 kJ
e. -150 kJ

Answer 1

The standard enthalpy of formation of RgO is nearest to -150 kJ (option e). It was calculated from the enthalpies of the reactions Rg₂O(s) + 1/2O₂(g) → 2RgO(s) (ΔH°₁ = -141 kJ) and Rg₂O(s) → Rg(s) + RgO(s) (ΔH°₂ = +11 kJ).    

The first balanced reaction is:

Rg₂O(s) + 1/2O₂(g) → 2RgO(s)   ΔH°₁ = -141 kJ    (1)  

The second balanced reaction is:

Rg₂O(s) → Rg(s) + RgO(s)  ΔH°₂ = +11 kJ   (2)  

Since we need to find the enthalpy of formation of RgO, we need to invert reaction (2), as follows:

Rg(s) + RgO(s) → Rg₂O(s)    ΔH°₂ = -11 kJ    (3)            

Now, after adding equations (1) and (3), we have:

Rg₂O(s) + 1/2O₂(g) + Rg(s) + RgO(s) → 2RgO(s) + Rg₂O(s)

Rg(s) + 1/2O₂(g) → RgO(s)     (4)  

The enthalpy of formation of RgO (reaction 4) is given by the sum of enthalpies ΔH°₁ and ΔH°₂, from equations (1) and (3), respectively

$\Delta H_{4}^{\circ} = -141 kJ + (-11 kJ) = -152 kJ$

Therefore, the standard enthalpy of formation of RgO is nearest to -150 kJ (option e).

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