madey21
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[tex]\[
2 K + 2 H_2O \rightarrow 2 KOH + H_2
\][/tex]

\(1.00 \, \text{g}\) of solid \(K\) is added to \(500 \, \text{mL}\) of water, initially at \(21.3^{\circ} \text{C}\). The temperature at the end of the reaction was \(23.5^{\circ} \text{C}\).

[tex]\[
c_{\text{soln}} = 4.18 \, \text{J/g} \cdot ^{\circ} \text{C} \quad d_{\text{soln}} = 1.00 \, \text{g/mL}
\][/tex]

What is the heat of reaction, [tex]\(q_{\text{rxn}}\)[/tex]?

Sagot :

GinnyAnswer
To solve this problem, we will go through the following steps:

1. Determine the mass of the solution:
- Here, 500 mL of water is provided, and since the density of the solution is 1.00 g/mL, we can calculate the mass of the solution.
- Mass of the solution \( = \text{volume of water} \times \text{density of the solution} \)
- Mass of the solution \( = 500 \text{ mL} \times 1.00 \text{ g/mL} = 500.0 \text{ g} \)

2. Calculate the change in temperature (\(\Delta T\)):
- Initial temperature \( T_i = 21.3^{\circ}C \)
- Final temperature \( T_f = 23.5^{\circ}C \)
- Change in temperature \( \Delta T = T_f - T_i \)
- \(\Delta T = 23.5^{\circ}C - 21.3^{\circ}C = 2.2^{\circ}C \)

3. Calculate the heat absorbed by the solution (\(q_{solution}\)):
- The heat absorbed by the solution can be calculated using the formula: \( q = mc\Delta T \)
- Where:
- \( m \) is the mass of the solution \( = 500.0 \text{ g} \)
- \( c \) is the specific heat capacity of the solution \( = 4.18 \text{ J/g}^{\circ}C \)
- \( \Delta T \) is the change in temperature \( = 2.2^{\circ}C \)
- So,
- \( q_{solution} = m \times c \times \Delta T \)
- \( q_{solution} = 500.0 \text{ g} \times 4.18 \text{ J/g}^{\circ}C \times 2.2^{\circ}C \)
- \( q_{solution} = 4598.0 \text{ J} \)

4. Determine the heat of reaction (\(q_{rxn}\)):
- The heat of reaction is the negative of the heat absorbed by the solution.
- Therefore, \( q_{rxn} = -q_{solution} \)
- \( q_{rxn} = -4598.0 \text{ J} \)

So, the heat of reaction, [tex]\( q_{rxn} \)[/tex], is [tex]\( -4598.0 \text{ J} \)[/tex].
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