Energetics

[aminzairi]

When 100 cm3 of 1.0 mol dm–3 HCl is mixed with 100 cm3 of 1.0 mol dm–3 NaOH, the temperature of the resulting solution increases by 5.0 °C. What will be the temperature change, in °C, when 50 cm3 of these two solutions are mixed? 

Can someone explain how to do this question ?

[kw0573]

Key idea: temperature is related to heat absorbed per mol of species (you learned this through calculating heat exchange using specific heat capacities), such heat in this case is associated with enthalpy of the acid-base reaction.

Two 50 cm³ solutions mixed yields an enthalpy half as much as that of mixing two 100 cm³ solutions of same concentration. The enthalpy is released as heat. Because there are only half as much mols of chemical species in the two 50 cm³ reaction, that also means half the heat spread to half the number of species results in the same temperature change of + 5.0°C.  

I assumed that the question meant mixing the same acid and base solution with 50 cm³ each, and not 50 cm³ in total. This was an IB multiple choice (paper 1) question.

[aminzairi]

3 minutes ago, kw0573 said:

Key idea: temperature is related to heat absorbed per mol of species (you learned this through calculating heat exchange using specific heat capacities), such heat in this case is associated with enthalpy of the acid-base reaction.

Two 50 cm³ solutions mixed yields an enthalpy half as much as that of mixing two 100 cm³ solutions of same concentration. The enthalpy is released as heat. Because there are only half as much mols of chemical species in the two 50 cm³ reaction, that also means half the heat spread to half the number of species results in the same temperature change of + 5.0°C.  

I assumed that the question meant mixing the same acid and base solution with 50 cm³ each, and not 50 cm³ in total. This was an IB multiple choice (paper 1) question.

so the change of temperature also depends on the number of molecule ? because the less the number of molecules that will absorb the enthalpy, the more the heat is released ? am i right ?

[kw0573]

11 hours ago, aminzairi said:

so the change of temperature also depends on the number of molecule ? 

Yes

 

11 hours ago, aminzairi said:

 because the less the number of molecules that will absorb the enthalpy, the more the heat is released ? am i right ?

No. There is an order issue. First we consider how much reactive molecules to make the reaction happen, which have a certain enthalpy change. More molecules that react will have a greater enthalpy change.

Now, enthalpy is usually in form of heat. Number of mols and specific heat capacities determine how the heat is distributed. The heat is distributed between all molecules, not just the products. 

[ZAYED573]

On 12/8/2016 at 2:45 PM, kw0573 said:

Key idea: temperature is related to heat absorbed per mol of species (you learned this through calculating heat exchange using specific heat capacities), such heat in this case is associated with enthalpy of the acid-base reaction.

Two 50 cm³ solutions mixed yields an enthalpy half as much as that of mixing two 100 cm³ solutions of same concentration. The enthalpy is released as heat. Because there are only half as much mols of chemical species in the two 50 cm³ reaction, that also means half the heat spread to half the number of species results in the same temperature change of + 5.0°C.  

I assumed that the question meant mixing the same acid and base solution with 50 cm³ each, and not 50 cm³ in total. This was an IB multiple choice (paper 1) question.

Bro, I'm so confused so as long as the cocnentration's the same but the volume is different, the answer is valid? I'm trying to use this info to answer another question:" When 25 cm3 of 1.0 mol dm-3 HCl was added to 25 cm3 of 1.0 mol dm-3 KOH the temperature increased by 7 oC. Which combination of HCl and KOH will also give a temperature rise of 7  degrees C?

Options

a) 50 cm3 of 0.25 mol dm-3 HCl and 50 cm3 of 0.25 mol dm-3 KOH.

b) 50 cm3 of 0. 50 mol dm-3 HCl and 50 cm3 of 0.50 mol dm-3 KOH.

c)50 cm3 of 1.0 mol dm-3 HCl and 50 cm3 of 1.0 mol dm-3 KOH.

d) 50 cm3 of 2.0 mol dm-3 HCl and 50 cm3 of 2.0 mol dm-3 KOH.

 

According to my teacher, the answer is C and all I understood is that by halving or doubling the volume of the same concentrations of bases and acids, you get the same rise in temperature.