Question

Calculate the heat of reaction Delta*H for the following reaction: C*H_{4}(g) + 2O_{2}(g) -> C*O_{2}(g) + 2H_{2}*O(g)

Answer 1

The heat of reaction (ΔH) for the combustion of methane (CH4) is -890.4 kJ, indicating an exothermic process where heat is released.

Step-by-step explanation:

Calculating Heat of Reaction using Hess's Law

To calculate the heat of reaction (ΔH) for the combustion of methane (CH4), where methane gas reacts with oxygen to produce carbon dioxide and water vapor, we will apply Hess's Law. The provided thermochemical equation is:

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) + 890.4 kJ

The positive sign of the ΔH value indicates that this amount of heat is produced or released during the reaction, suggesting that the process is exothermic.

Given the above equation, the heat of reaction for this process, which includes the transformation of methane and oxygen gas into carbon dioxide gas and water vapor, is -890.4 kJ. This value is negative to indicate an exothermic reaction where heat is a product of the chemical reaction.

Answer 2

The heat of reaction (ΔH) for the combustion of methane to produce gaseous water and carbon dioxide is calculated as 802.4 kJ by using Hess's Law and adjusting for the physical state of water in the reaction.

Step-by-step explanation:

To calculate the heat of reaction (ΔH) for the combustion of methane (CH4), we can use Hess's Law and thermochemical data provided. The initial reaction is CH4(g) + 2O2(g) → CO2(g) + 2H2O(g). Using the data given, heat is a product in the reaction: CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) + 890.4 kJ.

However, the physical state of water in the provided reaction is liquid (l) while in our desired reaction, water is in the gaseous state (g).

The enthalpy change will differ based on the physical state of water. We also have the information that when 1 mole of gaseous water forms, only 242 kJ of heat is released compared to 286 kJ for liquid water.

Therefore, to correct for the physical state, we need to adjust with the heat of vaporization of water which is the difference between the heat released for liquid water and gaseous water: 286 kJ - 242 kJ = 44 kJ per mole of water.

Since two moles of water are formed in the reaction, this amounts to 2 x 44 kJ = 88 kJ that should be subtracted from the 890.4 kJ provided for the liquid water.

The corrected ΔH for the formation of gaseous water is then 890.4 kJ - 88 kJ = 802.4 kJ.