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Answered

[tex]$19.64 \, \text{g} \, \text{H}_2\text{SO}_4$[/tex] dissociates 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]\[
\begin{aligned}
\text{H}_2\text{SO}_4 & \rightarrow \text{H}^+ + \text{HSO}_4^{-} \\
C_{\text{soln}} & = 3.50 \, \text{J} / \text{g} \, ^{\circ} \text{C} \\
q_{\text{rxn}} & = [?] \, \text{J}
\end{aligned}
\][/tex]

Enter either a + or - sign AND the magnitude.



Answer :

To determine the heat of the reaction ([tex]\( q_{\text{rxn}} \)[/tex]), we will follow a series of steps to calculate it based on the given data. Here’s a detailed, step-by-step solution:

1. Initial Data:

- Mass of [tex]\( \text{H}_2\text{SO}_4 \)[/tex]: [tex]\( 19.64 \)[/tex] grams
- Mass of water: [tex]\( 100.0 \)[/tex] grams
- Specific heat capacity of the solution ([tex]\( C_{\text{soln}} \)[/tex]): [tex]\( 3.50 \)[/tex] J/g°C
- Initial temperature: [tex]\( 23.12 \)[/tex] °C
- Final temperature: [tex]\( 57.30 \)[/tex] °C

2. Calculate Total Mass of the Solution:

The total mass of the solution is the sum of the mass of [tex]\( \text{H}_2\text{SO}_4 \)[/tex] and the mass of water:

[tex]\[ \text{Total mass of solution} = \text{mass}_{\text{H}_2\text{SO}_4} + \text{mass}_{\text{water}} \][/tex]

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

3. Calculate the Change in Temperature ([tex]\( \Delta T \)[/tex]):

[tex]\[ \Delta T = \text{final temperature} - \text{initial temperature} \][/tex]

[tex]\[ \Delta T = 57.30 \, °C - 23.12 \, °C = 34.18 \, °C \][/tex]

4. Calculate the Heat of the Reaction ([tex]\( q_{\text{rxn}} \)[/tex]) Using [tex]\( q = mc\Delta T \)[/tex]:

[tex]\[ q_{\text{rxn}} = \text{total mass of solution} \times \text{specific heat capacity of solution} \times \Delta T \][/tex]

[tex]\[ q_{\text{rxn}} = 119.64 \, \text{grams} \times 3.50 \, \text{J/g°C} \times 34.18 \, °C \][/tex]

[tex]\[ q_{\text{rxn}} = 119.64 \times 3.50 \times 34.18 = 14312.53 \, \text{J} \][/tex]

The heat of the reaction [tex]\( q_{\text{rxn}} \)[/tex] is positive because the temperature of the solution increased, indicating that the reaction is exothermic and heat was released into the surroundings.

Thus, the heat of the reaction is:

[tex]\[ +14312.53 \, \text{J} \][/tex]