Name: Spontaneity
Uploaded: 2019-07-19
Duration: 4 min 31 s
Description: What is the difference between a spontaneous process and a non-spontaneous process? Watch the video to understand the concept of spontaneity.

Question

Propyne undergoes complete combustion as follows:

$C_{3} H_{4} (g) + 4 O_{2} (g) \to 3 {CO}_{2} (g) + 2 H_{2} O (l)$

Calculate the enthalpy change of this reaction, given the following additional values:

$∆ H_{f}^{⊝}$ of ${CO}_{2} (g) = - 394 kJ {mol}^{- 1}$
$∆ H_{f}^{⊝}$ of $H_{2} O (l) = - 286 kJ {mol}^{- 1}$
$C_{3} H_{4} (g) + 4 O_{2} (g) \to 3 {CO}_{2} (g) + 2 H_{2} O (l)$ of $C_{3} H_{4} (g) = 185.4 kJ {mol}^{- 1}$

Accepted Answer

Propyne undergoes complete combustion as follows:

$C_{3} H_{4} (g) + 4 O_{2} (g) \to 3 {CO}_{2} (g) + 2 H_{2} O (l)$

The enthalpy change for a reaction is given as:
${ΔH}_{r} = {ΣΔH}_{f}^{⊝}$ (products) $- {ΣΔH}_{f}^{⊝}$ (reactants)

Therefore, the enthalpy change for the given reaction is $= 3 \times ∆ H_{f}^{⊝} ({CO}_{2}) + 2 \times ∆ H_{f}^{⊝} (H_{2} O) - ∆ H_{f}^{⊝} (C_{3} H_{4}) - ∆ H_{f}^{⊝} (O_{2}) = 3 \times (- 394) + 2 \times (- 286) - 185.4 - 0 = - 1939.4 kJ$

Note:- As oxygen ( $O_{2}$ ) exists in its standard state, the standard enthalpy of formation for oxygen gas is $0 kJ {mol}^{- 1}$ .

Important Questions on Energetics and Thermochemistry (AHL)

Learn and Prepare for any exam you want !

Propyne undergoes complete combustion as follows: C3H4(g) +4O2(g) → 3CO2(g) + 2H2O (l) Calculate the enthalpy change of this reaction, given the following additional values: ∆Hf⊝ of CO2(g)=-394 kJ mol-1 ∆Hf⊝ of H2O(l) = -286 kJ mol-1 ∆Hf⊝ of C3H4(g) = 185.4 kJ mol-1

Important Questions on Energetics and Thermochemistry (AHL)

Learn and Prepare for any exam you want !