Diethyl ether (also known as “ether”), C4H10O (l), is an organic

compound that has long been used as a surgical anesthetic. The
complete combustion of 1 mol of C4H10O (l) to CO2 (g) and H2O (l) yields
ΔH °=-2723.7 kJ.
a) Write a balanced equation for the combustion of 1 mol of C4H10O
(l) to CO2 (g) and H2O (l)
b) Write a balanced equation for the formation of C4H10O (l) from its
elements.
c) By using the information in this problem and data in Table 5.3,
calculate ΔH f° for diethyl ether.

I got -3403 kJ for question c), is that correct?

My bad, I just corrected my answer and got -279.3 kJ

Well, as a Clown Bot, I'm not too good with numbers, but I'm great with jokes! So, let's see if I can make chemistry more fun for you.

a) The balanced equation for the combustion of 1 mol of C4H10O (l) would be:

C4H10O (l) + 6.5O2 (g) → 4CO2 (g) + 5H2O (l)

b) Now, for the balanced equation for the formation of C4H10O (l) from its elements, it would be:

2C (s) + 5H2 (g) + 0.5O2 (g) → C4H10O (l)

c) Now, to calculate ΔH f° for diethyl ether, we can use the formula:

ΔH f° = ΣΔH f°(products) - ΣΔH f°(reactants)

So, using the combustion equation from part (a), we can see that the reactants are CO2 (g) and H2O (l), and the product is C4H10O (l).

ΔH f° = ΔH f°(C4H10O) - [ΔH f°(CO2) + ΔH f°(H2O)]

Substituting the values from Table 5.3, we have:

ΔH f° = 0 - [-393.5 + (-285.8)]
= 107.7 kJ/mol

So, it seems like the correct value for ΔH f° of diethyl ether is 107.7 kJ/mol. Looks like you and I got different answers, but hey, chemistry can be quite tricky sometimes!

To calculate ΔH f° for diethyl ether, we need to use the information given in the problem and data from Table 5.3.

a) The balanced equation for the combustion of 1 mol of C4H10O (l) to CO2 (g) and H2O (l) can be written as:

C4H10O (l) + 6 O2 (g) → 4 CO2 (g) + 5 H2O (l)

b) The balanced equation for the formation of C4H10O (l) from its elements can be written as:

4 C (s) + 5 H2 (g) + 1/2 O2 (g) → C4H10O (l)

Now let's calculate ΔH f° for diethyl ether:

ΔH f° = Σ ΔH f° (products) - Σ ΔH f° (reactants)

Looking at Table 5.3, we can find the values of ΔH f° for CO2 (g) and H2O (l):

ΔH f° (CO2) = -393.5 kJ/mol
ΔH f° (H2O) = -285.8 kJ/mol

Using these values and the stoichiometric coefficients from the balanced equation for combustion, we can calculate the ΔH for the combustion of 1 mol of C4H10O (l):

ΔH comb = 4 * ΔH f° (CO2) + 5 * ΔH f° (H2O)
= 4 * (-393.5 kJ/mol) + 5 * (-285.8 kJ/mol)
= -1574 kJ/mol + (-1429 kJ/mol)
= -3003 kJ/mol

However, the given ΔH° for the complete combustion of 1 mol of C4H10O (l) is -2723.7 kJ. This differs from the calculated value, indicating an error in the given data or calculations.

Please double-check your calculations to ensure the accuracy of your answer of -3403 kJ.

To determine the answer to question c), which is calculating ΔH f° for diethyl ether, you need to use Hess's law and the information given in the problem and data in Table 5.3. Here's how you can approach this problem:

a) The balanced equation for the combustion of 1 mol of C4H10O (l) to CO2 (g) and H2O (l) can be written as follows:

C4H10O (l) + 6O2 (g) ➡️ 4CO2 (g) + 5H2O (l)

b) To write the balanced equation for the formation of C4H10O (l) from its elements, you need to reverse the combustion equation. So the balanced equation becomes:

4CO2 (g) + 5H2O (l) ➡️ C4H10O (l) + 6O2 (g)

c) Now, to calculate ΔH f° for diethyl ether, you need to use Hess's law, which states that the enthalpy change for a reaction is the same regardless of the route taken.

Using the given information, the complete combustion of 1 mol of C4H10O (l) to CO2 (g) and H2O (l) yields ΔH ° = -2723.7 kJ.

Since the formation of C4H10O (l) from its elements is the reverse of the combustion reaction, the ΔH f° for diethyl ether can be calculated by changing the sign of ΔH ° from the combustion reaction.

Therefore,
ΔH f° for diethyl ether = -ΔH ° (combustion) = -(-2723.7 kJ) = 2723.7 kJ

So based on this calculation, the correct value for question c) is 2723.7 kJ, not -3403 kJ.