- #1

- 3

- 0

## Homework Statement

Calculate the heat of reaction per mole of water formed in the acid-base reaction. Assume that the total mass of the solution is 150g.

75mL of 2.0M NaOH added to 75mL of 2.0M HCl in a styrofoam cup. Average initial temp = 295.65K. final temp = 308.8K

## Homework Equations

Δ

*H*

_{rxn}= q

_{rxn}= -mcΔT

## The Attempt at a Solution

q

_{rxn}= -(141.234g)(4.186J/gK)(308.8K-295.65K) = -7.8kJ

based on: .15 mol of HCl (5.5g) and .15 mol NaOH (6.0g) reacted to form .15 mol H2O (2.7g) and .15 mol NaCl (8.8g) therefore since the entire reaction is aqueous and the total mass of the solution is 150g: 150g -8.8g (the amount of the product that is not water) = 141.2g

in a simpler train of thought i decided 2.70g of water were actually formed and therefore:

q

_{rxn}= -(2.70g)(4.186J/gK)(308.8K-295.65K) = -1.49*10

^{2}J

I feel that i am grossly over thinking this, however, and I need a bit of advice