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Sagot :
Sure, let's go through the solution step-by-step.
### Identifying the Reaction Type
First, we need to understand whether the reaction is endothermic or exothermic. The given reaction is:
[tex]\[ 2 \text{ NH}_3(g) \rightarrow \text{N}_2(g) + 3 \text{H}_2(g) \quad \Delta H = 160 \text{kJ} \][/tex]
Since the enthalpy change (ΔH) is positive (160 kJ), it means that the reaction absorbs heat from the surroundings. Hence, the reaction is endothermic.
### Determining the Number of Moles of NH₃
Next, we need to find out how many moles of NH₃ are present in 35.8 grams of NH₃. The molar mass of NH₃ (Ammonia) is calculated as follows:
[tex]\[ \text{Molar mass of NH}_3 = (1 \times 14.007) + (3 \times 1.008) = 17.031 \text{ g/mol} \][/tex]
To find the number of moles:
[tex]\[ \text{moles of NH}_3 = \frac{\text{mass of NH}_3}{\text{molar mass of NH}_3} = \frac{35.8 \text{ g}}{17.031 \text{ g/mol}} = 2.102 \text{ moles (rounded to 3 significant digits)} \][/tex]
### Calculating the Heat Absorbed
For the given reaction:
[tex]\[ 2 \text{ moles of NH}_3 \quad \to \quad \text{160 kJ}\][/tex]
We need to find out how much heat is absorbed when 2.102 moles of NH₃ react. Using the given data, we set up the proportions:
[tex]\[ \text{Heat absorbed} = \left( \frac{\text{moles of NH}_3}{2.0} \right) \times 160 \text{kJ} \][/tex]
[tex]\[ \text{Heat absorbed} = \left( \frac{2.102 \text{ moles}}{2.0} \right) \times 160 \text{kJ} = 168.164 \text{kJ} \][/tex]
Rounding this to 3 significant digits, we get:
[tex]\[ \text{Heat absorbed} = 168 \text{ kJ} \][/tex]
### Conclusion
Now we can fill in the answer for the given questions:
1. This reaction is...
- Endothermic.
2. Suppose 35.8 g of NH₃ react. Will any heat be released or absorbed?
- Yes, absorbed.
3. If you said heat will be released or absorbed in the second part of this question, calculate how much heat will be released or absorbed. Round your answer to 3 significant digits.
- 168 kJ.
So the formatted final answer for the given table would look like this:
\begin{tabular}{|c|c|}
\hline This reaction is... & \begin{tabular}{l}
endothermic. \\
\end{tabular} \\
\hline \begin{tabular}{l}
Suppose 35.8 g of [tex]$NH _3$[/tex] react. \\
Will any heat be released or absorbed?
\end{tabular} & \begin{tabular}{l}
Yes, absorbed. \\
\end{tabular} \\
\hline \begin{tabular}{l}
If you said heat will be released or absorbed in \\
the second part of this question, calculate how \\
much heat will be released or absorbed. \\
Round your answer to 3 significant digits.
\end{tabular} & 168 kJ \\
\hline
\end{tabular}
### Identifying the Reaction Type
First, we need to understand whether the reaction is endothermic or exothermic. The given reaction is:
[tex]\[ 2 \text{ NH}_3(g) \rightarrow \text{N}_2(g) + 3 \text{H}_2(g) \quad \Delta H = 160 \text{kJ} \][/tex]
Since the enthalpy change (ΔH) is positive (160 kJ), it means that the reaction absorbs heat from the surroundings. Hence, the reaction is endothermic.
### Determining the Number of Moles of NH₃
Next, we need to find out how many moles of NH₃ are present in 35.8 grams of NH₃. The molar mass of NH₃ (Ammonia) is calculated as follows:
[tex]\[ \text{Molar mass of NH}_3 = (1 \times 14.007) + (3 \times 1.008) = 17.031 \text{ g/mol} \][/tex]
To find the number of moles:
[tex]\[ \text{moles of NH}_3 = \frac{\text{mass of NH}_3}{\text{molar mass of NH}_3} = \frac{35.8 \text{ g}}{17.031 \text{ g/mol}} = 2.102 \text{ moles (rounded to 3 significant digits)} \][/tex]
### Calculating the Heat Absorbed
For the given reaction:
[tex]\[ 2 \text{ moles of NH}_3 \quad \to \quad \text{160 kJ}\][/tex]
We need to find out how much heat is absorbed when 2.102 moles of NH₃ react. Using the given data, we set up the proportions:
[tex]\[ \text{Heat absorbed} = \left( \frac{\text{moles of NH}_3}{2.0} \right) \times 160 \text{kJ} \][/tex]
[tex]\[ \text{Heat absorbed} = \left( \frac{2.102 \text{ moles}}{2.0} \right) \times 160 \text{kJ} = 168.164 \text{kJ} \][/tex]
Rounding this to 3 significant digits, we get:
[tex]\[ \text{Heat absorbed} = 168 \text{ kJ} \][/tex]
### Conclusion
Now we can fill in the answer for the given questions:
1. This reaction is...
- Endothermic.
2. Suppose 35.8 g of NH₃ react. Will any heat be released or absorbed?
- Yes, absorbed.
3. If you said heat will be released or absorbed in the second part of this question, calculate how much heat will be released or absorbed. Round your answer to 3 significant digits.
- 168 kJ.
So the formatted final answer for the given table would look like this:
\begin{tabular}{|c|c|}
\hline This reaction is... & \begin{tabular}{l}
endothermic. \\
\end{tabular} \\
\hline \begin{tabular}{l}
Suppose 35.8 g of [tex]$NH _3$[/tex] react. \\
Will any heat be released or absorbed?
\end{tabular} & \begin{tabular}{l}
Yes, absorbed. \\
\end{tabular} \\
\hline \begin{tabular}{l}
If you said heat will be released or absorbed in \\
the second part of this question, calculate how \\
much heat will be released or absorbed. \\
Round your answer to 3 significant digits.
\end{tabular} & 168 kJ \\
\hline
\end{tabular}
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