0 votes
31 views
ago in Chemical reactions by (350 points)
Here's a precise problem for a chemistry student to solve:Using Hess's Law, calculate the enthalpy change for the reaction where 2 moles of methane (CH4) reacts with 4 moles of oxygen gas (O2) to form 2 moles of carbon dioxide (CO2) and 4 moles of water (H2O), given the following known enthalpy changes:ΔH1 = -890 kJ/mol for the reaction: CH4 + 2O2 → CO2 + 2H2OΔH2 = -3

1 Answer

0 votes
ago by (610 points)
To solve this problem using Hess's Law, we need to manipulate the given reactions and their enthalpy changes to match the desired reaction. The desired reaction is:2 CH4 + 4 O2  2 CO2 + 4 H2OFirst, we need to break down the methane  CH4  into its constituent elements:CH4  C + 2 H2However, we don't have the enthalpy change for this reaction. We can obtain it by manipulating the given reactions.We can start by reversing the reaction H2:CO2  C + O2  H = +394 kJ/mol Next, we need to multiply the reaction H3 by 2 to get 2 moles of H2O:2  H2 +  1/2 O2  H2O   H = 2 * -286 kJ/mol = -572 kJ/mol Now, we can add the reversed H2 and the multiplied H3:CO2  C + O2  H = +394 kJ/mol 2  H2 +  1/2 O2  H2O   H = -572 kJ/mol -----------------------------------------------C + 2 H2 + O2  CO2 + 2 H2O  H = -178 kJ/mol This reaction represents the breakdown of methane into its constituent elements and their subsequent reaction with oxygen to form CO2 and H2O. Now, we can multiply this reaction by 2 to match the desired reaction:2  C + 2 H2 + O2  CO2 + 2 H2O   H = 2 * -178 kJ/mol = -356 kJ/mol Finally, we have the enthalpy change for the desired reaction:2 CH4 + 4 O2  2 CO2 + 4 H2O  H = -356 kJ/mol

Related questions

Welcome to Sarvan Science Q&A, where you can ask questions and receive answers from other members of the community.
...