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 the heat absorbed by the solution (\(q_{sol}\)) is \(q = mc\Delta T\), where \(m\) is the mass of the solution, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature.
Given \(m = 102\space g\), \(c = 4.20\space J/g^\circ C\), and \(\Delta T = 8.65^\circ C\).

Step2: Calculate \(q_{sol}\)

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

Step3: Determine \(q_{rxn}\)

We know that \(q_{rxn}=-q_{sol}\) (because the heat released by the reaction is absorbed by the solution, so they have opposite signs).
So \(q_{rxn}=- 3705.66\space J\) (we can round to an appropriate number of significant figures, but following the calculation, this is the value)

Answer:

\(-3710\) (or more precisely \(-3705.66\), depending on rounding. If we use the given values strictly: \(102\times4.20\times8.65 = 102\times36.33 = 3705.66\), so \(-3706\) or \(-3710\) when rounded appropriately)