madey21
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What is the enthalpy of the reaction, in [tex]kJ/mol[/tex], knowing that [tex]7.5 \, \text{g} \, \text{KNO}_3[/tex] absorbs [tex]2597 \, \text{J}[/tex] of energy?

[tex]\[
\begin{array}{l}
\text{KNO}_3 \rightarrow \text{K}^+ + \text{NO}_3^- \\
q_{\text{rxn}} = +2597 \, \text{J} \\
\Delta H_{\text{rxn}} = [?] \, \text{kJ/mol}
\end{array}
\][/tex]

Enter either a [tex]+[/tex] or [tex]-[/tex] sign and the magnitude. Use significant figures.


Sagot :

Certainly! To determine the enthalpy change (ΔH) of the reaction in kJ/mol given the data, we need to follow these steps:

### Step 1: Determine the molar mass of [tex]\( KNO_3 \)[/tex].

The molar mass of [tex]\( KNO_3 \)[/tex] is calculated by adding the atomic masses of potassium (K), nitrogen (N), and three oxygen (O) atoms:

[tex]\[ K = 39.1 \text{ g/mol}, \quad N = 14.0 \text{ g/mol}, \quad O_3 = 3 \times 16.0 \text{ g/mol} \][/tex]

[tex]\[ \text{Molar mass of } KNO_3 = 39.1 + 14.0 + 48.0 = 101.1 \text{ g/mol} \][/tex]

### Step 2: Convert the mass of [tex]\( KNO_3 \)[/tex] to moles.

Given mass of [tex]\( KNO_3 \)[/tex] is 7.5 g. To find the number of moles, we divide the mass by the molar mass:

[tex]\[ \text{Moles of } KNO_3 = \frac{7.5 \text{ g}}{101.1 \text{ g/mol}} = 0.074184 \text{ moles} \][/tex]

### Step 3: Convert the energy absorbed from Joules to kJ.

The energy absorbed by the reaction is given as 2597 J. Since 1 kJ = 1000 J, we convert Joules to kilojoules:

[tex]\[ \text{Energy absorbed} = \frac{2597 \text{ J}}{1000 \text{ J/kJ}} = 2.597 \text{ kJ} \][/tex]

### Step 4: Calculate the enthalpy change (ΔH) per mole of [tex]\( KNO_3 \)[/tex].

ΔH is the energy change per mole of substance. We divide the energy absorbed in kJ by the number of moles of [tex]\( KNO_3 \)[/tex]:

[tex]\[ \Delta H = \frac{2.597 \text{ kJ}}{0.074184 \text{ moles}} = 35.008 \text{ kJ/mol} \][/tex]

Since the reaction absorbs energy, ΔH will be positive:

[tex]\[ \Delta H = + 35.01 \text{ kJ/mol (to four significant figures)} \][/tex]

Therefore, the enthalpy change (ΔH) for the given reaction is [tex]\( +35.01 \, \text{kJ/mol} \)[/tex].
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