Question

hcl + naoh → nacl + h₂o
25.0 ml of 1.00 m hcl is added to
75.0 ml of 1.00 m naoh. the
resulting temperature change is an
increase of 8.65 °c.
$c_{sol} = 4.20 j/g °c$ $d_{sol} = 1.02 g/ml$
$mass_{soln} = 102 g$
what is the heat of reaction?
$q_{rxn} = ? j$
because the heat capacity of the calorimeter cannot be
determined, $q_{rxn} = -q_{sol}$.
enter either a + or - sign and the magnitude.

Explanation:

Step1: Recall the formula for heat absorbed by solution

The formula for heat (\(q\)) is \(q = mc\Delta T\), where \(m\) is mass, \(c\) is specific heat, and \(\Delta T\) is temperature change.
Given \(m = 102\space g\), \(c = 4.20\space J/g^\circ C\), \(\Delta T = 8.65^\circ C\).

Step2: Calculate \(q_{sol}\)

Substitute values into the formula:
\(q_{sol}=102\space g\times4.20\space J/g^\circ C\times8.65^\circ C\)
First, calculate \(102\times4.20 = 428.4\)
Then, \(428.4\times8.65 = 428.4\times(8 + 0.65)=428.4\times8+428.4\times0.65 = 3427.2+278.46 = 3705.66\space J\)

Step3: Relate \(q_{rxn}\) to \(q_{sol}\)

Given \(q_{rxn}=-q_{sol}\), so \(q_{rxn}=- 3705.66\space J\) (The reaction releases heat, so it's exothermic, hence negative sign)

Answer:

\(-3706\) (or \(-3705.66\))