2 Al(s) + 3 Cl2(g) → 2 AlCl3(s)
Into a cylinder with a moveable piston (pressure stays constant), a piece of aluminum is placed and chlorine gas is added. After they react, the temperature inside the piston is considerable higher than its initial temperature, but the volume inside the piston is consideraby smaller. What are the signs of q and w for the reaction?
Energy change is the sum of heat and work: ΔE = q + w. Work is calculated by: w = -PΔV
What is the change in energy (in joules) if a reaction absorbs 72.2 J of heat and increases in volume from 0.250L to 0.750L at a constant pressure of 0.993 atm? Please include the correct sign with your numerical result.
(Note: Make sure your units are consistent when combining energy terms. The conversion factor you need is: 1 L·atm = 101.3 J)
A reaction is done in a coffee-cup calorimeter by mixing equal volumes of reactants. If the initial temperature of the reactant solutions before then are mixed is 24.0 oC and the final temperature of the solution after the reaction has occurred is 22.2 oC, we can say that the reaction is __________ and the heat of reaction, ∆Hrxn , is written the a ___________ sign. For this reaction the enthalpy content of the products is ___________ than that of the reactants. Select the correct answers from the list below.
I'm gonna try the second one
delta V = 0.750 L- 0.250L
= 0.50 L
P= 0.933atm
w= -Pdelta V
w = - 0.993 atm *0.50 L
w= - 0.4965 atm *L
In Joules:
-0.4965 L* atm *101.3 J/ 1 L*atm= -50.31 joules
I think -50.31 joules of work is done (since energy is absored q is positive)
q is - since T is higher (exothermic).
w is - since work is done ON the system.
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.
For the reaction 2 Al(s) + 3 Cl2(g) → 2 AlCl3(s) in a cylinder with a movable piston (constant pressure), the signs of q and w can be determined as follows:
1. The temperature inside the piston is considerably higher after the reaction. This indicates that heat is being absorbed by the system. Hence, q (heat) is positive (+).
2. The volume inside the piston is considerably smaller after the reaction. This means work is being done on the system, compressing it. According to the equation w = -PΔV, since ΔV is negative (-) for a decrease in volume, w (work) will also be negative (-).
Therefore, the signs for q and w in this reaction are:
q = + (heat is absorbed)
w = - (work is done on the system)
Next, for the question about the change in energy:
ΔE = q + w
Given:
q = 72.2 J (absorbed heat)
ΔV = Vfinal - Vinitial = 0.750 L - 0.250 L = 0.500 L (change in volume)
P = 0.993 atm (constant pressure)
Using the equation w = -PΔV, we can calculate the work done:
w = - (0.993 atm) * (0.500 L) = -0.4965 L·atm
Since 1 L·atm = 101.3 J, converting the work to joules:
w = -0.4965 L·atm * (101.3 J / 1 L·atm) = -50.277 J
Now, substituting the known values into the energy change equation:
ΔE = q + w
ΔE = 72.2 J + (-50.277 J)
ΔE = 21.923 J
Therefore, the change in energy for this reaction is 21.923 J (with the correct sign being positive).
Moving on to the last question:
The initial temperature of the reactant solutions is 24.0 oC, while the final temperature of the solution after the reaction is 22.2 oC.
Based on these temperature changes, we can draw the following conclusions:
1. The reaction is exothermic because the final temperature is lower than the initial temperature. Heat is released during the reaction.
2. The heat of reaction (∆Hrxn) is negative (-) because heat is being released by the system.
3. The enthalpy content of the products is lower than that of the reactants because the reaction is exothermic, meaning that the products have less stored energy (enthalpy) than the reactants.
Therefore, the correct answers are:
- The reaction is exothermic.
- The heat of reaction (∆Hrxn) is written with a negative (-) sign.
- The enthalpy content of the products is lower than that of the reactants.