Question

agno₃ + hcl → agcl + hno₃
50.0 ml of 0.20 m agno₃ is added to
50.0 ml of 0.20 m hcl. the resulting
temperature change is an increase of
1.60 °c.
(c_{\text{sol}} = 4.20 , \text{j/g °c}) (quad d_{\text{sol}} = 1.02 , \text{g/ml}) (quad c_{\text{cal}} = 5.60 , \text{j/°c})
(\text{mass}_{\text{sol}} = 102 , \text{g}) (quad q_{\text{rxn}} = -694 , \text{j}) (quad 0.010 , \text{mol reacted})
what is the enthalpy of reaction?
(delta h_{\text{rxn}} = ? , \text{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 heat of the reaction ($q_{rxn}$) divided by the moles of the reactant that reacted. The formula is $\Delta H_{rxn}=\frac{q_{rxn}}{n}$, where $n$ is the moles of the reactant.

Step2: Identify the values of $q_{rxn}$ and $n$

We are given that $q_{rxn} = - 694\space J$ (the negative sign indicates that the reaction is exothermic as the temperature increased) and the moles of the reactant that reacted, $n = 0.010\space mol$.

Step3: Calculate $\Delta H_{rxn}$

Substitute the values into the formula:
$\Delta H_{rxn}=\frac{- 694\space J}{0.010\space mol}=- 69400\space J/mol$
Convert the units from $J/mol$ to $kJ/mol$ by dividing by 1000:
$\Delta H_{rxn}=\frac{- 69400\space J/mol}{1000}=- 69.4\space kJ/mol$

Answer:

-69.4