madey21
Answered

[tex]$19.64 \text{ g } H_2SO_4$[/tex] dissociated in [tex]$100.0 \text{ g}$[/tex] of water in a coffee cup calorimeter. The temperature rose from [tex]$23.12^{\circ} \text{C}$[/tex] to [tex]$57.30^{\circ} \text{C}$[/tex]. What is the heat of the reaction, [tex]$q_{\text{rxn}}$[/tex]?

[tex]\[
\begin{array}{c}
H_2SO_4 \rightarrow H^{+} + HSO_4^{-} \\
c_{\text{soln}} = 3.50 \text{ J/g}^{\circ}\text{C} \\
q_{\text{rxn}} = [?] \text{ J}
\end{array}
\][/tex]



Answer :

To determine the heat of the reaction ([tex]\( q_{\text{rxn}} \)[/tex]), we follow a series of steps to calculate the heat absorbed by the solution upon the dissociation of sulfuric acid ([tex]\( H_2SO_4 \)[/tex]) in water, considering the given information. Here's a detailed step-by-step solution:

### Step 1: Calculate the Total Mass of the Solution

The total mass of the solution is the sum of the mass of [tex]\( H_2SO_4 \)[/tex] and the mass of water.
Given:
- Mass of [tex]\( H_2SO_4 \)[/tex] = [tex]\( 19.64 \)[/tex] grams
- Mass of water = [tex]\( 100.0 \)[/tex] grams

[tex]\[ \text{Total mass} = 19.64 \, \text{grams} + 100.0 \, \text{grams} = 119.64 \, \text{grams} \][/tex]

### Step 2: Calculate the Change in Temperature ([tex]\( \Delta T \)[/tex])

The change in temperature ([tex]\( \Delta T \)[/tex]) is the difference between the final temperature and the initial temperature.
Given:
- Initial temperature = [tex]\( 23.12 \, ^\circ \text{C} \)[/tex]
- Final temperature = [tex]\( 57.30 \, ^\circ \text{C} \)[/tex]

[tex]\[ \Delta T = 57.30 \, ^\circ \text{C} - 23.12 \, ^\circ \text{C} = 34.18 \, ^\circ \text{C} \][/tex]

### Step 3: Calculate the Heat of the Reaction ([tex]\( q_{\text{rxn}} \)[/tex])

To find [tex]\( q_{\text{rxn}} \)[/tex], we use the formula for heat absorbed or released by a solution:

[tex]\[ q_{\text{rxn}} = \text{Total mass} \times c_{\text{soln}} \times \Delta T \][/tex]

Where:
- [tex]\( \text{Total mass} = 119.64 \, \text{grams} \)[/tex]
- [tex]\( c_{\text{soln}} \)[/tex] (specific heat capacity of the solution) = [tex]\( 3.50 \, \text{J/g} \cdot ^\circ \text{C} \)[/tex]
- [tex]\( \Delta T = 34.18 \, ^\circ \text{C} \)[/tex]

Substitute the values into the equation:
[tex]\[ q_{\text{rxn}} = 119.64 \, \text{grams} \times 3.50 \, \text{J/g} \cdot ^\circ \text{C} \times 34.18 \, ^\circ \text{C} \][/tex]

### Step 4: Calculation

Multiplying these values together we get:
[tex]\[ q_{\text{rxn}} \approx 119.64 \times 3.50 \times 34.18 \][/tex]

[tex]\[ q_{\text{rxn}} \approx 14312.53 \, \text{J} \][/tex]

### Final Answer

The heat of the reaction ([tex]\( q_{\text{rxn}} \)[/tex]) is:
[tex]\[ q_{\text{rxn}} \approx 14312.53 \, \text{J} \][/tex]