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.
$c_{sol} = 4.20 j/g °c$ $d_{sol} = 1.05 g/ml$ $c_{cal} = 4.50 j/°c$
$mass_{sol} = 116 g$
what is the heat of reaction?
remember: $q_{rxn} = -(q_{sol} + q_{cal})$
$q_{rxn} = ? j$
enter either a + or - sign and the magnitude.

Explanation:

Step1: Calculate \( q_{\text{sol}} \)

The formula for heat absorbed by the solution is \( q_{\text{sol}} = c_{\text{sol}} \times m_{\text{sol}} \times \Delta T \). We know \( c_{\text{sol}} = 4.20 \, \text{J/g}^\circ\text{C} \), \( m_{\text{sol}} = 116 \, \text{g} \), and \( \Delta T = 1.20^\circ\text{C} \).
\( q_{\text{sol}} = 4.20 \, \text{J/g}^\circ\text{C} \times 116 \, \text{g} \times 1.20^\circ\text{C} \)
\( q_{\text{sol}} = 4.20 \times 116 \times 1.20 \, \text{J} \)
\( q_{\text{sol}} = 574.56 \, \text{J} \)

Step2: Calculate \( q_{\text{cal}} \)

The formula for heat absorbed by the calorimeter is \( q_{\text{cal}} = c_{\text{cal}} \times \Delta T \). We know \( c_{\text{cal}} = 4.50 \, \text{J/}^\circ\text{C} \) and \( \Delta T = 1.20^\circ\text{C} \).
\( q_{\text{cal}} = 4.50 \, \text{J/}^\circ\text{C} \times 1.20^\circ\text{C} \)
\( q_{\text{cal}} = 5.40 \, \text{J} \)

Step3: Calculate \( q_{\text{rxn}} \)

Using the formula \( q_{\text{rxn}} = -(q_{\text{sol}} + q_{\text{cal}}) \). First, find the sum of \( q_{\text{sol}} \) and \( q_{\text{cal}} \): \( 574.56 + 5.40 = 579.96 \, \text{J} \). Then, apply the negative sign: \( q_{\text{rxn}} = - 579.96 \, \text{J} \approx -580 \, \text{J} \) (rounded to a reasonable number of significant figures, or keeping the precision from calculation).

Answer:

\(-580\) (or more precisely \(-579.96\))