Question

Consider these reactions were in represents a generic metal one is to M solid +6 HCl aqueous healing to MCL three aqueous Blessed three H2 gas the delta H of one is -811.0 kJ reaction number two is HCl gas hearing HCl aqueous delta eight of two is -74.8 kJ problem number three H2 gas plus CO2 gas healing to HCl gas Delta age of three is -1845.0 kJ number for MCL three solid building MCL three aqueous the Delta 84 is -445.0kj use the given information to determine the enthalpy of the reaction 2 M(s) + 3 Cl2(g) yielding 2MCl3(s)

Answer 1

The enthalpy of the reaction 2 M(s) + 3 Cl2(g) → 2MCl3(s) is found by using Hess's law to rearrange and sum the enthalpies of given reactions, ensuring stoichiometric coefficients align with the desired reaction.

Step-by-step explanation:

To determine the enthalpy of the reaction 2 M(s) + 3 Cl2(g) → 2MCl3(s), we must use Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for individual steps that lead to the overall reaction.

The given reactions are:

1. M(s) + 6 HCl(aq) → MCl3(aq) + 3 H2(g), ΔH = -811.0 kJ
2. HCl(g) → HCl(aq), ΔH = -74.8 kJ
3. H2(g) + Cl2(g) → 2 HCl(g), ΔH = -1845.0 kJ
4. MCl3(solid) → MCl3(aqueous), ΔH = -445.0kJ

First, we need to adjust the reactions to match the desired reaction. This might involve reversing and multiplying through by appropriate stoichiometric coefficients. Then we add the enthalpies for each step, taking into account the changes made.