madey21
Answered

Solve for \( x \).

[tex]\[ 3x = 6x - 2 \][/tex]



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$
2 K +2 H _2 O \rightarrow 2 KOH + H _2
$
[tex]$1.00 g$[/tex] of solid [tex]$K$[/tex] is added to [tex]$500 . mL$[/tex] of water initially at [tex]$21.3^{\circ} C$[/tex]. The temperature at the end of the reaction was [tex]$23.5^{\circ} C$[/tex].
$
c _{\text {soln }}=4.18 J / g ^{\circ} C \quad d _{\text {soln }}=1.00 g / mL
$

What is the heat of reaction, [tex]$q_{r \times n}$[/tex] ?
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Response:
Given the reaction:

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

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

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



Answer :

GinnyAnswer
Let's work through this problem step by step.

### Given Data:
1. Mass of solid \( K \) = \( 1.00 \) gram
2. Volume of water = \( 500 \) mL
3. Initial temperature of the water = \( 21.3^\circ C \)
4. Final temperature of the solution after reaction = \( 23.5^\circ C \)
5. Specific heat capacity of the solution, \( c_{\text{soln}} \) = \( 4.18 \text{ J/g}^\circ \text{C} \)
6. Density of the solution, \( d_{\text{soln}} \) = \( 1.00 \text{ g/mL} \)

### Steps for Calculation:

#### Step 1: Calculate the mass of the solution
The mass of the solution can be calculated using the volume of water and the density of the solution.

Since the density \( d_{\text{soln}} \) is \( 1.00 \text{ g/mL} \), the mass of the solution is:
[tex]\[ \text{mass}_{\text{solution}} = \text{volume}_{\text{water}} \times \text{density}_{\text{soln}} = 500.0 \text{ mL} \times 1.0 \text{ g/mL} = 500.0 \text{ grams} \][/tex]

#### Step 2: Calculate the change in temperature (\(\Delta T\))
The change in temperature (\(\Delta T\)) is given by the final temperature minus the initial temperature.

[tex]\[ \Delta T = 23.5^\circ C - 21.3^\circ C = 2.2^\circ C \][/tex]

#### Step 3: Calculate the heat absorbed by the solution (\(q\))
The heat absorbed by the solution (\(q\)) can be calculated using the formula:
[tex]\[ q = m \times c_{\text{soln}} \times \Delta T \][/tex]
where:
- \( m \) is the mass of the solution
- \( c_{\text{soln}} \) is the specific heat capacity of the solution
- \( \Delta T \) is the change in temperature

Substitute the known values into the formula:
[tex]\[ q = 500.0 \text{ grams} \times 4.18 \text{ J/g}^\circ \text{C} \times 2.2^\circ C \][/tex]

#### Step 4: Calculate the numerical value of heat (\(q\))
[tex]\[ q = 500.0 \times 4.18 \times 2.2 \approx 4598 \text{ J} \][/tex]

### Conclusion
The heat of reaction, [tex]\( q_{r \times n} \)[/tex], associated with the dissolution and reaction of [tex]\( 1.00 \)[/tex] gram of [tex]\( K \)[/tex] in [tex]\( 500 \)[/tex] mL of water, resulting in the temperature change from [tex]\( 21.3^\circ C \)[/tex] to [tex]\( 23.5^\circ C \)[/tex], is approximately [tex]\( 4598 \text{ J} \)[/tex].