[helenee]

In an experiment, a student mixes a 50 mL sample of 0.100 M AgNO3 with a 50 mL sample of 0.1 NaCl at 20°C in a coffee cup calorimeter. What's the enthalpy change of the precipitation reaction represented here if the temperature final is 21°C? (The total mass of the mixture is 100 grams and specific heat capacity is 4.2 J/g°C).

AgNO3 + NaCl --> agCl + NaNO3

I did solve the problem and got the magnitude right: 84 kJ/mol.

However, I didn't put the - (negative) sign and got the answer wrong.

The problem is, I don't see why it has to be negative.

Please help me understand this!

[Borek]

Was the reaction exo-, or endothermic? What does the convention say about sign of the ΔH for the exothermic (endothermic) reactions?

[helenee]

Since the final temperature is 21°C and increased (from 20), the reaction must have released more heat and hence was exothermic (am I right?). Since exothermic reactions have negative delta H, is that why the answer is negative?

[JoeB]

That's correct; the equation to calculate the heat energy (the equation we use is q=mcΔT, though the exact symbols you use may be slightly different) will always give you a positive answer, so when you then go on to calculate the enthalpy change (ΔH) you need to take another look at the question and decide whether or not the reaction is endothermic or exothermic.

If it is endothermic and the temperature decreases then you can just leave it how it is as a positive answer, but (like in this case) the reaction is exothermic since the temperature increases, you have to adjust your answer at the end to be negative.

[Borek]

Since the final temperature is 21°C and increased (from 20), the reaction must have released more heat and hence was exothermic (am I right?). Since exothermic reactions have negative delta H, is that why the answer is negative?

Yes.

[Borek]

the equation to calculate the heat energy (the equation we use is q=mcΔT, though the exact symbols you use may be slightly different) will always give you a positive answer

No. It all depends on how you calculate ΔT. If you define it as

[tex]\Delta T = T_{final} - T_{initial}[/tex]

(and use this definition consistently) calculated ΔH will be either positive or negative (depending on the exact values of the temperature before and after the reaction).