The balanced equation for the formation reaction of butane, C4H10, is:

4C(s, graphite) + 5H2(g) → C4H10(l)

To calculate the enthalpy change of formation, ∆𝑓𝐻298𝐾𝜃 𝐶4𝐻10(𝑙), we can use Hess's Law, which states that the total enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps into which a reaction can be divided.

The enthalpy change of formation of butane can be calculated from the given reactions and their enthalpies:

1. C4H8(l) + 6O2(g) → 4CO2(g) + 4H2O(l) ∆𝑐𝑜𝑚𝑏.𝐻298𝐾𝜃 = -2,718.8 kJ mol-1
2. C4H8(l) + H2(g) → C4H10(l) ∆ℎ𝑦𝑑.𝐻298𝐾𝜃 = -126.8 kJ mol-1
3. C(s, graphite) + O2(g) → CO2(g) ∆𝑓𝐻298𝐾𝜃 (𝐶𝑂2(𝑔)) = -393.5 kJ mol-1
4. H2(g) + 12 O2(g) → H2O(l) ∆𝑓𝐻298𝐾𝜃 (𝐻2𝑂(𝑙)) = -285.8 kJ mol-1

We can write the formation of butane as the sum of these reactions:

4C(s, graphite) + 5H2(g) → C4H10(l) ∆𝑓𝐻298𝐾𝜃 𝐶4𝐻10(𝑙) = ?

We can rearrange the given reactions to match the formation reaction of butane:

1. Multiply reaction 3 by 4 and reverse it: 4CO2(g) → 4C(s, graphite) + 4O2(g) ∆𝑓𝐻298𝐾𝜃 = 4 * 393.5 kJ mol-1 = 1574 kJ mol-1
2. Multiply reaction 4 by 5: 5H2(g) + 5/2 O2(g) → 5H2O(l) ∆𝑓𝐻298𝐾𝜃 = 5 * -285.8 kJ mol-1 = -1429 kJ mol-1
3. Add reaction 2 as it is: C4H8(l) + H2(g) → C4H10(l) ∆ℎ𝑦𝑑.𝐻298𝐾𝜃 = -126.8 kJ mol-1
4. Reverse reaction 1 and subtract the enthalpy change of reaction 2 from it: C4H10(l) → C4H8(l) + 6O2(g) ∆𝑐𝑜𝑚𝑏.𝐻298𝐾𝜃 = 2718.8 kJ mol-1 - (-126.8 kJ mol-1) = 2845.6 kJ mol-1

Adding these reactions gives the formation reaction of butane, and the enthalpy change of formation is the sum of the enthalpy changes of these reactions:

∆𝑓𝐻298𝐾𝜃 𝐶4𝐻10(𝑙) = 1574 kJ mol-1 - 1429 kJ mol-1 - 126.8 kJ mol-1 + 2845.6 kJ mol-1 = 1863.8 kJ mol-1

So, the enthalpy change of formation of butane is 1863.8 kJ mol-1.

Question

The balanced equation for the formation reaction of butane, C4H10, is:

4C(s, graphite) + 5H2(g) → C4H10(l)

To calculate the enthalpy change of formation, ∆𝑓𝐻298𝐾𝜃 𝐶4𝐻10(𝑙), we can use Hess's Law, which states that the total enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps into which a reaction can be divided.

The enthalpy change of formation of butane can be calculated from the given reactions and their enthalpies:

1. C4H8(l) + 6O2(g) → 4CO2(g) + 4H2O(l) ∆𝑐𝑜𝑚𝑏.𝐻298𝐾𝜃 = -2,718.8 kJ mol-1
2. C4H8(l) + H2(g) → C4H10(l) ∆ℎ𝑦𝑑.𝐻298𝐾𝜃 = -126.8 kJ mol-1
3. C(s, graphite) + O2(g) → CO2(g) ∆𝑓𝐻298𝐾𝜃 (𝐶𝑂2(𝑔)) = -393.5 kJ mol-1
4. H2(g) + 12 O2(g) → H2O(l) ∆𝑓𝐻298𝐾𝜃 (𝐻2𝑂(𝑙)) = -285.8 kJ mol-1

We can write the formation of butane as the sum of these reactions:

4C(s, graphite) + 5H2(g) → C4H10(l) ∆𝑓𝐻298𝐾𝜃 𝐶4𝐻10(𝑙) = ?

We can rearrange the given reactions to match the formation reaction of butane:

1. Multiply reaction 3 by 4 and reverse it: 4CO2(g) → 4C(s, graphite) + 4O2(g) ∆𝑓𝐻298𝐾𝜃 = 4 * 393.5 kJ mol-1 = 1574 kJ mol-1
2. Multiply reaction 4 by 5: 5H2(g) + 5/2 O2(g) → 5H2O(l) ∆𝑓𝐻298𝐾𝜃 = 5 * -285.8 kJ mol-1 = -1429 kJ mol-1
3. Add reaction 2 as it is: C4H8(l) + H2(g) → C4H10(l) ∆ℎ𝑦𝑑.𝐻298𝐾𝜃 = -126.8 kJ mol-1
4. Reverse reaction 1 and subtract the enthalpy change of reaction 2 from it: C4H10(l) → C4H8(l) + 6O2(g) ∆𝑐𝑜𝑚𝑏.𝐻298𝐾𝜃 = 2718.8 kJ mol-1 - (-126.8 kJ mol-1) = 2845.6 kJ mol-1

Adding these reactions gives the formation reaction of butane, and the enthalpy change of formation is the sum of the enthalpy changes of these reactions:

∆𝑓𝐻298𝐾𝜃 𝐶4𝐻10(𝑙) = 1574 kJ mol-1 - 1429 kJ mol-1 - 126.8 kJ mol-1 + 2845.6 kJ mol-1 = 1863.8 kJ mol-1

So, the enthalpy change of formation of butane is 1863.8 kJ mol-1.

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