The diagrams show gases that are stored in two separate but similar containers.
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Answer:
0.018 moles of isooctane must be burned to produce 100 kJ.
Explanation:
When the isooctane is <em>burned</em>, it undergoes a reaction known as combustion, in which reacts with oxygen to produce carbon dioxide and water. In standard conditions (Pressure = 1 atm and Temperature = 298 K) formed water is liquid. The balanced equation is:
C₈H₁₈(l) + 12.5 O₂(g) ⇄ 8 CO₂(g) + 9 H₂O(l)
The standard enthalpy of combustion (which can be found in tables) relates the amount of isooctane that reacts and the heat the reaction liberates. In this case, ΔH°c = -5,460 kJ/mol, that is every 1 mol of isooctane that burns, 5,460 kJ of heat are released.
We can begin with the information of 100 kJ and apply this conversion factor. Since this energy is released, by convention, we write it with a minus sign.
To calculate this, we will use the chemical equations as math equations and add them.
Firtly, we want the equation for the formation of CH₃CHO(g), so this will be the only product.
The reactants must be only the elements in their standard form, so C(g), O₂(g) and H₂(g). I would be more correct to use C(s), but since we odn't have information for this, we will assume it wants with C(g).
So, the reaction we want is:
To balance the reaction, we can just do for eqach element separately, maintaining the coefficient of 1 on CH₃CHO(g):
Now, we want to get to this equation adding the equations we want. We will apply the same operations to the enthalpies to get the enthalpy of formation.
The first given equation has the CH₃CHO(g), but it is on the left side and with coefficient of 2, so we need to invert the reaction and divided every coefficient by 2. The same operations have to be applied to the enthalpy, so the sign of the enthalpy will invert and it will be divided by 2:
The second given equation has both C(g) and O₂(g), but since the third equation also has O₂(g), we will look just for C(g). We need 2 C(g), so we will need to doulbe the equation and its enthalpy:
For the last, we will look into H₂(g) and since all the equations are balanced, O₂(g) will also be balanced by the end of it.
We need 2 H₂(g), so we don't need to do anything with this reaction:
Now, we add the equations:
And we do the same with the enthalpies:
This is the enthalpy for this reaction. To get the molar enthalpy of formation, we need to divide this value by the coefficient of CH₃CHO(g). Since this coefficient is 1, we have:
So, the molar enthalpy of formation given the data is -271.2 kJ/mol.