Answer:
Approximately 241 kJ/mol reaction.
Explanation:
How many moles of the reaction took place?
The coefficient in front of both NaOH and HCl are both one. In other words, one mole of the reaction will require one mole of NaOH and one mole and HCl.
[tex]\rm0.025\; L \times 0.100\; mol\cdot L^{-1} = 0.0025\; mol[/tex] of both NaOH and HCl are available. As a result, 0.0025 moles of the reaction would have taken place.
How much heat was produced?
Approximate volume of the solution:
[tex]\rm 25.0 + 25.0 = 50.0\; mL[/tex].
Approximate mass of the solution:
[tex]\rm 50.0\; mL\times 1\;g\cdot mL^{-1} = 50.0\; g[/tex].
[tex]Q = c\cdot m \cdot \Delta T = \rm 50.0\; g \times 4.812\; J \cdot g^{-1} \cdot K^{-1} \times (25.5 - 23.0)\; K = 601.5\; J[/tex]
Enthalpy change per mole reaction:
[tex]\displaystyle \Delta H = \frac{Q}{n} = \rm \frac{601.5\; J}{0.025\; mol} \approx 241000\; J\cdot mol^{-1} = 241\; kJ \cdot mol^{-1}[/tex].
