Question

Consider the following two reactions: A → 2B ΔH°rxn = 456.7 kJ/mol A → C ΔH°rxn = -22.1 kJ/mol Determine the enthalpy change for the process: 2B → C Consider the following two reactions: A 2B H°rxn = 456.7 kJ/mol A C H°rxn = -22.1 kJ/mol Determine the enthalpy change for the process: 2B C

Answer 1

The enthalpy change for the process 2B → C can be determined using Hess's Law. By adding the enthalpy changes of the given reactions, we find that the enthalpy change for 2B → C is 891.3 kJ/mol.

Step-by-step explanation:

The enthalpy change for the process 2B → C can be determined using the concept of Hess's Law. We can use the given reactions to find the desired reaction. By adding the enthalpy changes of the given reactions, we can determine the enthalpy change for the desired reaction. The enthalpy change for the process 2B → C is the sum of the enthalpy change for the reaction A → C and twice the enthalpy change for the reaction A → 2B.

Given: A → 2B ΔH°rxn = 456.7 kJ/mol and A → C ΔH°rxn = -22.1 kJ/mol.

Substituting the given values, we have: 2B → C ΔH°rxn = (2 × 456.7 kJ/mol) + (-22.1 kJ/mol) = 891.3 kJ/mol.

Answer 2

To find the enthalpy change for 2B → C, we apply Hess's Law by reversing the first reaction A → 2B and combining it with the second reaction A → C. This gives a total enthalpy change of -478.8 kJ for the 2B → C process.

Step-by-step explanation:

To calculate the enthalpy change for the process 2B → C, we need to use the concept of Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps into which a reaction can be divided. The first given reaction is A → 2B with an enthalpy change (ΔH°rxn) of +456.7 kJ/mol. The second reaction is A → C with an enthalpy change of -22.1 kJ/mol.

We know that enthalpy is a state function, meaning that it does not depend on the path taken from reactants to products, only the initial and final states. So, to get from 2B → C, we can reverse the first reaction to get B from A, and then use the second reaction to get to C from A. When we reverse the first reaction, we change the sign of the enthalpy, making it -456.7 kJ/mol for A → 2B. Adding it to the enthalpy for the second reaction, we get:

(-456.7 kJ/mol) + (-22.1 kJ/mol) = -478.8 kJ/mol for 2B → C.

The standard enthalpy change for the reaction 2B → C is therefore ΔH° = -478.8 kJ.