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_soln = 116 g q_rxn = -584 j 0.024 moles_rxn
what is the enthalpy of reaction?
δh_rxn = ? kj/mol
enter either a + or - sign and the magnitude.

Explanation:

Step1: Recall Enthalpy Formula

Enthalpy of reaction ($\Delta H_{rxn}$) is given by $\Delta H_{rxn} = \frac{q_{rxn}}{n_{rxn}}$, where $q_{rxn}$ is heat of reaction and $n_{rxn}$ is moles of reaction.

Step2: Substitute Values

Given $q_{rxn} = -584\ J = -0.584\ kJ$ (convert J to kJ: divide by 1000) and $n_{rxn} = 0.024\ mol$.
So, $\Delta H_{rxn} = \frac{-0.584\ kJ}{0.024\ mol}$

Step3: Calculate the Value

$\frac{-0.584}{0.024} \approx -24.33\ kJ/mol$ (rounded appropriately)

Answer:

-24.3 kJ/mol (or more precise -24.33 kJ/mol, depending on rounding)