Question

2naoh + h₂so₄ → 2h₂o + na₂so₄
50.0 ml of 0.50 m h₂so₄ is added to
60.0 ml of 0.80 m naoh. the resulting
temperature change is an increase of
1.20 °c.
$c_{sol}$ = 4.20 j/g °c $d_{sol}$ = 1.05 g/ml $c_{cal}$ = 4.50 j/°c
$mass_{soln}$ = 116 g $q_{rxn}$ = −584 j 0.024 $moles_{rxn}$
what is the enthalpy of reaction?
$\delta h_{rxn}$ = ? kj/mol
enter either a + or - sign and the magnitude.

Explanation:

Step1: Recall the formula for enthalpy of reaction

The enthalpy of reaction (\(\Delta H_{rxn}\)) is given by the heat of the reaction (\(q_{rxn}\)) divided by the moles of the reaction (\(n_{rxn}\)), i.e., \(\Delta H_{rxn}=\frac{q_{rxn}}{n_{rxn}}\).

Step2: Substitute the given values

We know that \(q_{rxn} = - 584\space J=-0.584\space kJ\) (since \(1\space kJ = 1000\space J\)) and \(n_{rxn}=0.024\space mol\).

Substituting these values into the formula: \(\Delta H_{rxn}=\frac{- 0.584\space kJ}{0.024\space mol}\)

Step3: Calculate the value

\(\frac{-0.584}{0.024}\approx - 24.33\space kJ/mol\) (rounded to two decimal places)

Answer:

\(-24.33\) (or more precisely, if we calculate \(\frac{-584\space J}{0.024\space mol}=\frac{-0.584\space kJ}{0.024\space mol}\approx - 24.33\space kJ/mol\))