Question

When a student mixes 50 of M HCI and 50 mL of 1.0 M NaoH in a coffee-cup calorimeter, the temperature of the resultant solution increases from 21.0 degree C to 27.5 degree C. Calculate the enthal change (delta H) for the reaction in kJ/mol HCl. (Assuming that the calorimeter loses only a negligible quantity heat, that the total volume of the solution is 100 ml, that its density is 1.0 g/mL, and that its specific heat is 4.18 J/g middot K.)

Answer 1

(II)

Let's start by writing the neutralization reaction in case we need it later:

msi = 100mL / lg 1ml Csolz 4. 18-1 sol= 2717丁 mol HCl 50 mL HCi = 0.05 mol HCİ 000 mL HCi -24仔J 0.05mol Hci

HCl + NaOH -> H₂O + NaCl

Now, we will use the calorimeter/specific heat formula:

q = mcΔT

We have c = 4.18J/gK and we also have ΔT=6.5K, we need to calculate m (but we have volume and density) so m=100g.

Now we just substitute the values in the formula and we get a q = 2717J but that isn't the final answer because we didn't use 1 mole of HCl, so we need to calculate how many moles we used. In this case we used 0.05moles HCl. So to get ΔH we need to divide q/moles (be very careful to include the negative sign in q because this heat is released).

So, the final answer is ΔH = - 54.34 kJ/mol