Question

h₂so₄ + 2koh → k₂so₄ + 2h₂o
50.0 ml of 2.00 m h₂so₄ is added to
50.0 ml of 2.00 m koh. the resulting
temperature change is an increase of
14.5 °c.
$c_{sol}$ = 4.20 j/g °c $d_{sol}$ = 1.00 g/ml $c_{cal}$ = 5.70 j/°c
$mass_{soln}$ = 100. g $q_{rxn}$ = -6, 170 j 0.0500 moles$_{rxn}$
what is the enthalpy of reaction?
$\delta h_{rxn}$ = ? 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: Identify the values of \(q_{rxn}\) and \(n_{rxn}\)

From the problem, we have \(q_{rxn}=- 6170\space J\) (the negative sign indicates that the reaction is exothermic as the temperature increased) and \(n_{rxn} = 0.0500\space mol\).

Step3: Convert \(q_{rxn}\) to kJ

We know that \(1\space kJ=1000\space J\), so to convert \(q_{rxn}\) from joules to kilojoules, we divide by 1000.
\(q_{rxn}=\frac{- 6170\space J}{1000}=-6.17\space kJ\)

Step4: Calculate \(\Delta H_{rxn}\)

Using the formula \(\Delta H_{rxn}=\frac{q_{rxn}}{n_{rxn}}\), substitute \(q_{rxn}=- 6.17\space kJ\) and \(n_{rxn}=0.0500\space mol\)
\(\Delta H_{rxn}=\frac{-6.17\space kJ}{0.0500\space mol}=- 123.4\space kJ/mol\) (We can also verify using the heat from the solution and the calorimeter. The heat absorbed by the solution \(q_{sol}=m_{sol}c_{sol}\Delta T + c_{cal}\Delta T\). Let's check: \(m_{sol} = 100\space g\), \(c_{sol}=4.20\space J/g^{\circ}C\), \(\Delta T = 14.5^{\circ}C\), \(c_{cal}=5.70\space J/^{\circ}C\). \(q_{sol}=100\times4.20\times14.5+5.70\times14.5=100\times60.9 + 82.65=6090+82.65 = 6172.65\space J\approx6170\space J\). Since the reaction is exothermic, \(q_{rxn}=-q_{sol}=- 6170\space J\). Then \(\Delta H_{rxn}=\frac{q_{rxn}}{n_{rxn}}=\frac{-6170\space J}{0.0500\space mol}=\frac{-6.17\space kJ}{0.0500\space mol}=- 123.4\space kJ/mol\approx - 123\space kJ/mol\) (or more accurately - 123.4 kJ/mol))

Answer:

\(-123\) (or \(-123.4\))