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.

c_sol = 4.20 j/g °c d_sol = 1.00 g/ml
mass_sol = 55.0 g q_rxn = -2,210 j 0.0250 moles_rxn

what is the enthalpy of reaction?
δh_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 formula $\Delta H_{rxn}=\frac{q_{rxn}}{n_{rxn}}$, where $q_{rxn}$ is the heat of the reaction and $n_{rxn}$ is the moles of the reaction.

Step2: Substitute the given values

We are given that $q_{rxn} = - 2210\ J=-2.210\ kJ$ (since $1\ kJ = 1000\ J$) and $n_{rxn}=0.0250\ mol$.
Substituting these values into the formula: $\Delta H_{rxn}=\frac{- 2.210\ kJ}{0.0250\ mol}$

Step3: Calculate the value

$\frac{-2.210}{0.0250}=- 88.4\ kJ/mol$

Answer:

-88.4