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 q_rxn = -3,710 j 0.0250 mol reacted
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 heat of the reaction ($q_{rxn}$) divided by the moles of the reactant that reacted ($n$), i.e., $\Delta H_{rxn}=\frac{q_{rxn}}{n}$.

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

We are given that $q_{rxn} = - 3710\ J$ and $n=0.0250\ mol$.

Step3: Convert $q_{rxn}$ to kJ

Since $1\ kJ = 1000\ J$, we convert $q_{rxn}$: $q_{rxn}=-3710\ J\times\frac{1\ kJ}{1000\ J}=-3.710\ kJ$.

Step4: Calculate $\Delta H_{rxn}$

Using the formula $\Delta H_{rxn}=\frac{q_{rxn}}{n}$, substitute $q_{rxn}=-3.710\ kJ$ and $n = 0.0250\ mol$:
$\Delta H_{rxn}=\frac{-3.710\ kJ}{0.0250\ mol}=-148.4\ kJ/mol$ (we can also check with the given data, alternatively we can think about the heat released by the reaction, the temperature increased so the reaction is exothermic, so $\Delta H$ should be negative. The calculation: $\frac{- 3710\ J}{0.0250\ mol}=\frac{-3.710\ kJ}{0.0250\ mol}=-148.4\ kJ/mol$, which can be rounded to - 148 kJ/mol or more accurately - 148.4 kJ/mol. But let's do the calculation precisely: $3710\div0.0250 = 148400\ J/mol=148.4\ kJ/mol$, and since the reaction is exothermic (releases heat, $q_{rxn}$ is negative), $\Delta H_{rxn}$ is negative. So $\Delta H_{rxn}=\frac{-3710\ J}{0.0250\ mol}=-148400\ J/mol=-148.4\ kJ/mol$)

Answer:

-148.4 (or -148 if rounded, but the precise calculation gives -148.4)