Consider the reaction:
M + 2HCl �¨ MCl2 + H2
When 0.85 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 6795 J of heat. What is ∆H in kJ per mol of M for this reaction? (Hint: Is this reaction exothermic or endothermic?)
The reaction is exothermic and you know that because the temperature of the water increased.
q = 6795 J for 0.85 mol; therefore,
q = 6795/0.85 = dH = ?
Remember to add a - sign since this is exothermic. Also remember to convert from J to kJ.
-7.99
4.7
Well, this reaction is as hot as an angry chili pepper! It's definitely exothermic because the temperature of the solution rose. Now, to calculate ∆H in kJ per mol of M, we can use the heat produced in the reaction.
Given that 0.85 mol of M reacted and produced 6795 J of heat, let's convert it to kJ. Remember, kilo means "thousand," so we divide by 1000.
6795 J ÷ 1000 = 6.795 kJ
So, ∆H for this reaction is 6.795 kJ per mol of M.
To calculate the enthalpy change (∆H) in kJ per mol of M for this reaction, we need to use the heat (q) produced and the moles of M reacted.
The heat released or absorbed in a reaction can be calculated using the equation:
q = ∆H * n
where q is the heat produced or absorbed, ∆H is the enthalpy change, and n is the number of moles of the substance involved in the reaction.
Given that the reaction produced 6795 J of heat and the number of moles of M reacted is 0.85 mol, we can rearrange the equation to solve for ∆H:
∆H = q / n
∆H = 6795 J / 0.85 mol
Now, let's convert the answer from J to kJ by dividing it by 1000:
∆H = (6795 J / 0.85 mol) / 1000 = 7.99 kJ/mol
Therefore, the enthalpy change (∆H) for this reaction is approximately 7.99 kJ per mol of M.
Now, let's determine if the reaction is exothermic or endothermic.
Since heat is produced in this reaction (∆H is negative), it means the reaction is exothermic.