Question

a chemist measures the energy change \\( \delta h \\) during the following reaction:

\\( 2 \mathrm { hgo } ( s ) \
ightarrow 2 \mathrm { hg } ( l ) + \mathrm { o } _ { 2 } ( g ) \quad \delta h = 182. \mathrm { kj } \\)

use the information to answer the following questions.

this reaction is...	\\( \bigcirc \\) endothermic. \\( \bigcirc \\) exothermic.
if you said heat will be released or absorbed in the second part of this question, calculate how much heat will be released or absorbed. be sure your answer has the correct number of significant digits.	\\( \square \mathrm { kj } \\)

Explanation:

First Question: Determine if the reaction is endothermic or exothermic

In a chemical reaction, if the enthalpy change (\(\Delta H\)) is positive, the reaction absorbs heat from the surroundings, which means it is endothermic. Here, \(\Delta H = 182\space kJ\) (positive), so the reaction is endothermic.

Since the reaction is endothermic (as determined above), when HgO reacts, heat will be absorbed. So for 31.4 g of HgO reacting, heat will be absorbed.

Step 1: Molar mass of HgO

The molar mass of Hg (mercury) is approximately \(200.59\space g/mol\) and the molar mass of O (oxygen) is \(16.00\space g/mol\). So the molar mass of \(HgO\) is \(200.59 + 16.00 = 216.59\space g/mol\).

Step 2: Moles of HgO

Given mass of HgO is \(31.4\space g\). Moles of HgO, \(n=\frac{mass}{molar\space mass}=\frac{31.4\space g}{216.59\space g/mol}\approx0.14497\space mol\).

Step 3: Moles of reaction

From the balanced equation \(2\space mol\) of \(HgO\) reacts with \(\Delta H = 182\space kJ\) (absorbed). So for \(2\space mol\) HgO, heat absorbed is \(182\space kJ\). For \(n\) moles of HgO, the heat absorbed (\(q\)) is calculated using the ratio. The ratio of moles of HgO in the reaction is \(\frac{n}{2}\) (since 2 moles of HgO correspond to 182 kJ). So \(q=\frac{n\times182\space kJ}{2}\).
Substituting \(n = 0.14497\space mol\), we get \(q=\frac{0.14497\space mol\times182\space kJ}{2}\).

Step 4: Calculate the heat

First, calculate \(0.14497\times182 = 26.38454\), then divide by 2: \(\frac{26.38454}{2}=13.19227\space kJ\). Rounding to the correct number of significant digits (3 significant digits from 31.4 g), the heat absorbed is approximately \(13.2\space kJ\).

Answer:

endothermic.

Second Question: Heat release or absorption for 31.4 g HgO reaction