Question

A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO3 in a coffee cup calorimeter. If both solutions were initially at 35.00°C and the temperature of the resulting solution was recorded as 37.00°C, determine the ΔH°rxn (in units of kJ/mol NaOH) for the neutralization reaction between aqueous NaOH and HCl. The specific heat capacity of water = 4.18 J/g°C. Assume 1) that no heat is lost to the calorimeter or the surroundings, and 2) that the density and the heat capacity of the resulting solution are the same as water.

Answer 1

Answer:

ΔH = -55.73 kJ/mol

Explanation:

Since the density of water is 1 g/ml and the problem wants us to use this value to find the mass of NaOH and HNO3 we arrange the original equation of

$d = \frac{m}{v}\\m = d X v\\\\m = 100g (1*100)$

Since no heat was lost or gained

qrxn+qsoln=0. ----> qrxn=-qsoln.

qsol = mc*($t_{f} - t_{i}$)

The mass of the solution is the mass of NaOh and HNO3. 100+100= 200.

You plug in the remaining numbers using 4.18 as the heat capacity and 2 as the change in temperature:

qsol=200*(4.18)*(2)= 1672 J

qrxn= -1672 J

Now  ΔH = qrxn over the amount in moles or grams. This problem asks for NaOH in moles. To calculate this we use the concentration formula

M= $\frac{n}{V}$. ----->  n=V*M.

Note that you must convert 100 ml to L to use this equation. So we have

n=0.1*0.300 where n=0.03

ΔH =  $\frac{-1672}{0.03}$

ΔH = -55733.3 J/mol

ΔH = -55.73 kJ/mol