madey21
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You previously determined that 0.050 mol [tex]$BaCl_2$[/tex] formed during the reaction. What is the enthalpy of the reaction?

[tex]
\begin{array}{c}
Ba(OH)_2 + 2HCl \rightarrow BaCl_2 + 2H_2O \\
q_{\text{soln}} = 5934 \, \text{J}
\end{array}
[/tex]



Answer :

To determine the enthalpy of the reaction, let's follow a step-by-step approach:

1. Given Data:
- The number of moles of [tex]\( \text{BaCl}_2 \)[/tex] formed is [tex]\( 0.050 \)[/tex] mol.
- The amount of energy released in the solution ([tex]\( q_\text{soln} \)[/tex]) is [tex]\( 5934 \)[/tex] Joules.

2. Understanding the Concept:
- The enthalpy of the reaction ([tex]\( \Delta H \)[/tex]) represents the amount of energy change associated with the reaction. This is typically given in Joules per mole (J/mol).
- We know how much energy ([tex]\( q_\text{soln} \)[/tex]) is released for a known quantity of substance (moles of [tex]\( \text{BaCl}_2 \)[/tex]).

3. Calculation of Enthalpy Change:
- To find the enthalpy change per mole of [tex]\( \text{BaCl}_2 \)[/tex] formed, we will divide the total energy released by the number of moles formed.

[tex]\[ \Delta H = \frac{q_\text{soln}}{\text{moles of } \text{BaCl}_2} \][/tex]

4. Substitute the Given Values:
- Given [tex]\( q_\text{soln} = 5934 \)[/tex] J and moles of [tex]\( \text{BaCl}_2 = 0.050 \)[/tex] mol,

[tex]\[ \Delta H = \frac{5934 \text{ J}}{0.050 \text{ mol}} \][/tex]

5. Perform the Division:
[tex]\[ \Delta H = 118680 \text{ J/mol} \][/tex]

6. Interpret the Result:
- The positive value indicates the energy release per mole, meaning the reaction is exothermic, releasing energy to the surroundings.

Conclusion:
The enthalpy of the reaction, when [tex]\( \text{Ba(OH)}_2 \)[/tex] reacts with [tex]\( \text{HCl} \)[/tex] to form [tex]\( \text{BaCl}_2 \)[/tex] and water, is [tex]\( 118680 \)[/tex] Joules per mole of [tex]\( \text{BaCl}_2 \)[/tex] formed.

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