Answer:
The standard enthalpy of formation of HgO is -90.7 kJ/mol.
Explanation:
The reaction between Hg and oxygen is as follows.
From the given,
Molar mass of HgO = 216.59 g/mol
Mass of HgO decomposed = 18.5 g
Amount of heat absorbed = 7.75 kJ
From the reaction,
The standard enthalpy of formation = 
During the decomposition of 1 mol of HgO , 90.7 kJ of energy absorbed.
For the formation of 1 mol of HgO , 90.7 kJ of energy is release
Therefore, the enthalpy of formation of mercury(II)Oxide is -90.7 kJ/mol
The correct answer for the question that is being presented above is this one: "C) The reaction is spontaneous at high temperatures."
Since dH is positive and dS ~ 0 then -TdS ~ 0, meaning dG will always be greater than 0 so t<span>he reaction is spontaneous at high temperatures.</span>
<h3>
Answer:</h3>
The Equilibrium would shift to produce more NO
<h3>
Explanation:</h3>
The reaction is;
N₂(g) + O₂(g) ⇆ 2NO(g)
- When a reaction is at equilibrium then the forward reaction rate will be equivalent to the reverse reaction rate. Additionally, the concentration of the reactants and products are the same.
- From Le Chatelier's principle, additional reactants favor the formation of more products while additional products favor the formation of more reactants.
- For example, when more oxygen is added then more Nitrogen (II) oxide will be formed.
- Oxygen is a reactant and when increased it favors forward reaction which leads to the formation of more NO which is the product.
Kc= (nh4)
--------
(nh3) + (h2o)
