Answer :
The H value for the reaction when 66.9 g Hg is reacted with oxygen is 30.3 kJ. The correct answer is option C.
To determine the enthalpy change (ΔH) for the reaction, we need the balanced chemical equation for the reaction of mercury (Hg) with oxygen (O₂).
The balanced equation for the reaction is:
2Hg + O₂ → 2HgO
Now, we can calculate the ΔH value using the given mass of mercury (Hg) and the molar enthalpy of the formation of mercury(II) oxide (HgO). The molar enthalpy of the formation of HgO is -90.8 kJ/mol.
First, we need to calculate the number of moles of Hg:
Number of moles of Hg = mass of Hg / molar mass of Hg
The molar mass of Hg is 200.59 g/mol.
Number of moles of Hg = 66.9 g / 200.59 g/mol ≈ 0.333 mol
According to the balanced equation, 2 moles of Hg react to form 2 moles of HgO. Therefore, the reaction involving 0.333 mol of Hg will produce 0.333 mol of HgO.
Now, we can calculate the ΔH value:
ΔH = moles of product × molar enthalpy of formation
ΔH = 0.333 mol × (-90.8 kJ/mol) ≈ -30.3 kJ
The negative sign indicates that the reaction is exothermic.
Therefore, the correct answer is: C) -30.3 kJ
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