Question

1. what is the standard molar enthalpy of combustion for the following reaction

2 c₂h₂(g) + 5 o₂(g) → 4 co₂(g) + 2 h₂o(l) + 2.50×10³ kj

Explanation:

Step1: Analyze the given reaction

The given reaction is \(2C_2H_2(g)+5O_2(g)\to4CO_2(g) + 2H_2O(l)+2.50\times10^{3}\space kJ\). The enthalpy change \(\Delta H\) for the reaction as written is \(- 2.50\times10^{3}\space kJ\) (negative because heat is released in combustion).

Step2: Calculate molar enthalpy of combustion

The molar enthalpy of combustion (\(\Delta H_{c}^{\circ}\)) is the enthalpy change per mole of the substance combusted. Here, \(n = 2\space mol\) of \(C_2H_2(g)\) reacts.

We use the formula \(\Delta H_{c}^{\circ}=\frac{\Delta H}{n}\)

Substitute \(\Delta H=- 2.50\times10^{3}\space kJ\) and \(n = 2\space mol\)

\(\Delta H_{c}^{\circ}=\frac{-2.50\times 10^{3}\space kJ}{2\space mol}=-1250\space kJ/mol\)

Answer:

The standard molar enthalpy of combustion of \(C_2H_2(g)\) is \(-1250\space kJ/mol\)