For the first question:
The reaction 2 Al(s) + 3 Cl2(g) → 2 AlCl3(s) involves the formation of aluminum chloride solid from aluminum metal and chlorine gas. Given that the temperature inside the piston after the reaction is considerably higher and the volume is considerably smaller, we can determine the signs of q (heat) and w (work) as follows:
- The increase in temperature indicates that heat is absorbed by the system (the reaction is endothermic), so the sign of q is positive (+).
- The decrease in volume indicates that work is done on the system (the piston is compressed), so the sign of w is negative (-).
Therefore, the signs of q and w for the reaction are +q and -w.
For the second question:
To calculate the change in energy (ΔE) of the reaction, we need to use the formula ΔE = q + w. Given that the work (w) is calculated as w = -PΔV, where P is the constant pressure and ΔV is the change in volume, we can calculate ΔE as follows:
ΔE = q + w
ΔE = 72.2 J + (-PΔV)
ΔE = 72.2 J + (-0.993 atm * (0.750 L - 0.250 L) * 101.3 J/L·atm)
By substituting the values, we can calculate the numerical result. However, we need to ensure that the pressure unit (atm) and the volume unit (L) are consistent with the energy unit (J). Therefore, we'll convert 1 L·atm to J using the conversion factor provided:
1 L·atm = 101.3 J
ΔE = 72.2 J + (-0.993 atm * (0.500 L) * 101.3 J)
ΔE = 72.2 J + (-50.6 J)
ΔE = 21.6 J
Therefore, the change in energy of the reaction is 21.6 J (with the correct sign).
For the third question:
Given that the initial temperature of the reactant solutions is higher (24.0°C) than the final temperature after the reaction (22.2°C), we can conclude that the reaction is exothermic. This is because the system loses heat to the surroundings, resulting in a decrease in temperature.
Based on the principles of thermodynamics, an exothermic reaction releases heat energy, so the heat of reaction (∆Hrxn) is written with a negative sign (-). This indicates that the reaction releases energy to the surroundings.
Additionally, since the enthalpy (heat content) of the products is lower than that of the reactants, we can say that the enthalpy change (∆Hrxn) is negative (-). This suggests that the products have lower energy content than the reactants.