Question

Consider the reaction: M + 2HCl → MCl2 + H2

When 0.25 mol of the metal, M, reacted with an aqueous HCl solution (the HCl is in excess), the temperature of the solution rose because the reaction produced 7025 J of heat. What is ∆H in kJ per mol of M for this reaction? (Hint: Is this reaction exothermic or endothermic?)

Answer 1

Answer: The enthalpy change of the reaction is -28.1 kJ/mol

Step-by-step explanation:

To calculate the enthalpy change of the reaction, we use the equation:

`\Delta H_(rxn)=(q)/(n)`

where,

`q` = amount of heat released = -7025 J

n = number of moles of metal M = 0.25 moles

`\Delta H_(rxn)` = enthalpy change of the reaction

Putting values in above equation, we get:

`\Delta H_(rxn)=(-7025J)/(0.25mol)=-28100J/mol=-28.1kJ/mol`

Conversion factor used: 1 kJ = 1000 J

Sign convention of heat:

• When heat is absorbed, the sign of heat is taken to be positive and is written on the reactant side and is considered as an endothermic reaction.
• When heat is released, the sign of heat is taken to be negative and is written on the product side and is considered as an exothermic reaction.

So, the chemical reaction becomes:

`M+2HCl\rightarrow MCl_2+H_2+28.1kJ`

Hence, the enthalpy change of the reaction is -28.1 kJ/mol