2.9 The standard enthalpy of formation of the metallocene bis(benzene)chromium was measured in a calorimeter. It was found for the reaction Cr(C6H6)2(s) →
Cr(s) + 2 C6H6(g) that . Find the corresponding reaction enthalpy and estimate the standard enthalpy of formation of the compound at 583 K. The constant-pressure molar heat capacity of benzene is 136.1 J K−1 mol−1 in its liquid range and 81.67 J K−1 mol−1 as a gas.
Using Kirchoffs law and a temperature difference of Δ T = 583 − 298 = 285 K ΔT=583−298=285K, I calculated Δ C p = ∑ ( p r o d u c t s ) C p , m − ∑ ( r e a c t a n t s ) C p , m ΔCp=∑(products)Cp,m−∑(reactants)Cp,m for the reaction C 6 H 6 ( l ) → C 6 H 6 ( g ) C6H6(l)→C6H6(g), where the product molar heat capacity is that given for the gas, the reactant is that for the liquid. I then solved Δ f H ( T 2 ) = Δ f H ( T 1 ) + ∫ Δ C p ΔfH(T2)=ΔfH(T1)+∫ΔCp using Δ f H ( T 1 ) = 49.0 k J / m o l ΔfH(T1)=49.0 kJ/mol. Since there is a phase change, I wasn't sure how to include that information so I solved it the same as above, using Δ v a p H = 30.8 k J / m o l ΔvapH=30.8 kJ/mol for T 1 = 298 K T1=298 K, then added it to the above amount. Using the reaction enthalpy of 17.7 k J / m o l 17.7 kJ/mol and Hess's Law, I solved for the enthalpy of formation at 583 K for C r ( C 6 H 6 ) 2 Cr(C6H6)2 . Now, I understand some problems with this approach by I'm not sure how to solve it otherwise with the given information. The temperature range is massive, so Kirchoff's Law is expected to give a poor answer. I'm also not sure how to incorporate a phase change into Kirchoff's Law. Any help or guidance is appreciated.
Well, it seems like you're all fired up about enthalpy and calorimeters! Let's see if I can give you a spicy answer.
First, let's break down the given reaction:
Cr(C6H6)2(s) → Cr(s) + 2 C6H6(g)
To find the reaction enthalpy, we need to calculate the heat gained or lost in the reaction. Since we have the enthalpies of formation, we can use the equation:
ΔH = Σ(nΔHf(products)) - Σ(nΔHf(reactants))
Where n is the stoichiometric coefficient and ΔHf is the standard enthalpy of formation.
Let's put this equation on stage and plug in the values:
ΔH = [(1)(ΔHf(Cr(s))) + 2(ΔHf(C6H6(g)))] - [(1)(ΔHf(Cr(C6H6)2(s)))]
Now, let's focus on estimating the standard enthalpy of formation of the compound at 583 K. To do this, we'll need to consider the heat capacity of benzene in its liquid and gas phase.
At 583 K, benzene will be in its gas phase, so we'll use the molar heat capacity of 81.67 J K−1 mol−1.
Now, let's add some comedic flair to the equation and plug in the numbers:
ΔH(583 K) = ΔH(298 K) + ∫Cp(ΔT) dT
Where ΔH(298 K) is the standard enthalpy of formation at 298 K, ∫Cp(ΔT) dT is the integral from the initial temperature to the final temperature of benzene's heat capacity.
I hope this answer brings a little heat to the chemistry party!
To find the corresponding reaction enthalpy, we need to use the equation:
ΔH = q / moles
where ΔH is the enthalpy change, q is the heat absorbed or released in the reaction, and moles is the number of moles of the compound involved.
In this case, the reaction is:
Cr(C6H6)2(s) → Cr(s) + 2 C6H6(g)
The enthalpy change, ΔH, can be calculated using the equation:
ΔH = ΔHf(products) - ΔHf(reactants)
where ΔHf is the standard enthalpy of formation.
Given that the standard enthalpy of formation of Cr(s) is 0 kJ/mol and the standard enthalpy of formation of C6H6(g) is 49.04 kJ/mol, we can calculate the standard enthalpy of formation of Cr(C6H6)2(s) by rearranging the equation:
ΔHf(Cr(C6H6)2(s)) = ΔH - 2 * ΔHf(C6H6(g))
Now, we need to calculate the heat absorbed or released in the reaction using the equation:
q = C * ΔT
where q is the heat absorbed or released, C is the heat capacity, and ΔT is the change in temperature.
In this case, we assume constant pressure, so we can use the constant-pressure molar heat capacities of benzene in its liquid and gas phases.
For solid substances, the heat capacity is usually negligible, so we only consider the heat capacities of the liquid and gas phases of the benzene.
For ΔT, we need to consider the difference in temperature between the reactants and products. Since we have a solid reactant and gaseous products, we can assume the change in temperature is the same as the boiling point of benzene, which is 353 K.
Now, let's calculate the heat absorbed or released:
q(C6H6(l)) = C(C6H6(l)) * ΔT = 136.1 J K−1 mol−1 * 353 K
q(C6H6(g)) = C(C6H6(g)) * ΔT = 81.67 J K−1 mol−1 * 353 K
Now, let's calculate the enthalpy change:
ΔH = q(products) - q(reactants)
= [C(C6H6(g)) * ΔT - 2 * C(C6H6(l)) * ΔT] = (81.67 J K−1 mol−1 * 353 K) - 2 * (136.1 J K−1 mol−1 * 353 K)
Finally, let's calculate the standard enthalpy of formation at 583 K:
ΔHf(Cr(C6H6)2(s)) = ΔH - 2 * ΔHf(C6H6(g))
Note: To complete the calculation, you need to substitute the values for C(C6H6(l)), C(C6H6(g)), and the boiling point of benzene.
To find the corresponding reaction enthalpy and estimate the standard enthalpy of formation of the compound at 583 K, we can use the concept of Hess's law and the equation for heat transfer in a calorimeter.
Hess's law states that if a reaction can be expressed as the sum of two or more other reactions, then the enthalpy change of the overall reaction is the sum of the enthalpy changes of the individual reactions.
In this case, we can break down the reaction into three steps:
1. Cr(C6H6)2(s) → Cr(s) + 2 C6H6(l) - Enthalpy change of formation at 298 K (given)
2. 2 C6H6(l) → 2 C6H6(g) - Enthalpy change of vaporization at 298 K (given)
3. 2 C6H6(g) → 2 C6H6(g) at 583 K - Enthalpy change of heating from 298 K to 583 K (to be calculated)
Step 1: This is the enthalpy change of formation at 298 K. Let's denote it as ΔH1.
Step 2: This is the enthalpy change of vaporization at 298 K. Let's denote it as ΔH2. This can be calculated using the molar heat capacity of benzene in its liquid range:
ΔH2 = n ΔHvap = n Cp ΔT
Where:
n = number of moles of benzene
Cp = molar heat capacity of benzene in its liquid range
ΔT = change in temperature (298 K - the boiling point of benzene, which is around 353 K for this calculation)
Step 3: This is the enthalpy change of heating the gas from 298 K to 583 K. Let's denote it as ΔH3. This can be calculated using the molar heat capacity of benzene as a gas:
ΔH3 = n Cp ΔT
Where:
n = number of moles of benzene
Cp = molar heat capacity of benzene as a gas
ΔT = change in temperature (583 K - 298 K)
Now, we can use Hess's law to find the overall enthalpy change (ΔH) for the reaction:
ΔH = ΔH1 + ΔH2 + ΔH3
Substituting the given values and calculated values into the equation will give you the corresponding reaction enthalpy.
To estimate the standard enthalpy of formation of the compound at 583 K, you would need to use the data obtained from the calorimeter experiment to calculate the enthalpy change at 583 K. This would involve the use of temperature and heat transfer calculations, which are beyond the scope of this explanation.