Question

h₂so₄ + 2koh → k₂so₄ + 2h₂o
30.0 ml of 1.30 m h₂so₄ is added to
25.0 ml of 2.00 m koh. the resulting
temperature change is an increase of
9.55 °c.

cₛₒₗ = 4.20 j/g °c dₛₒₗ = 1.00 g/ml
massₛₒₗ = 55.0 g
what is the heat of reaction?
qᵣₓₙ = ? j
because the heat capacity of the calorimeter cannot be
determined, qᵣₓₙ = −qₛₒₗ.
enter either a + or - sign and the magnitude.

Explanation:

Step1: Recall the formula for heat absorbed by solution

The formula for heat (\(q\)) is \(q = c \times m \times \Delta T\), where \(c\) is specific heat, \(m\) is mass, and \(\Delta T\) is temperature change.
Given \(c_{sol}=4.20\ J/g\ ^\circ C\), \(m_{sol}=55.0\ g\), \(\Delta T = 9.55\ ^\circ C\).

Step2: Calculate \(q_{sol}\)

Substitute values into the formula:
\(q_{sol}=4.20\ J/g\ ^\circ C\times55.0\ g\times9.55\ ^\circ C\)
First, calculate \(4.20\times55.0 = 231\), then \(231\times9.55 = 2206.05\ J\).

Step3: Find \(q_{rxn}\) using \(q_{rxn}=-q_{sol}\)

Since the reaction releases heat (temperature increases, exothermic), \(q_{rxn}\) is negative.
\(q_{rxn}=-2206.05\ J\) (or approximately \(-2210\ J\) when rounded appropriately).

Answer:

\(-2210\) (or \(-2206\) if more precise)