At Westonci.ca, we connect you with the answers you need, thanks to our active and informed community. Get detailed answers to your questions from a community of experts dedicated to providing accurate information. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
Sure, let's go through the problem step-by-step.
1. First, we need to understand the reaction provided:
[tex]\[ N_2H_4(l) + O_2(g) \rightarrow N_2(g) + 2H_2O(g) \][/tex]
2. The standard enthalpy change ([tex]\(\Delta H^{\circ}\)[/tex]) for this reaction is given as:
[tex]\[ \Delta H^{\circ} = -543.0 \ \frac{kJ}{mol} \][/tex]
This means that for every mole of [tex]\(N_2H_4\)[/tex] reacting with [tex]\(O_2\)[/tex] under standard conditions, 543.0 kJ of energy is released (hence the negative sign).
3. Since the given problem states that the reaction remains unchanged to connect with the goal reaction, the enthalpy change for the modified reaction is the same as the standard enthalpy change provided.
4. Therefore, the enthalpy for the modified reaction is:
[tex]\[ \Delta H^{\circ} = -543.0 \ \frac{kJ}{mol} \][/tex]
This concludes the detailed explanation. The final answer should include the sign and the magnitude of the enthalpy change with an appropriate unit and significant figures.
[tex]\[ \boxed{-543.0 \ \frac{kJ}{mol}} \][/tex]
1. First, we need to understand the reaction provided:
[tex]\[ N_2H_4(l) + O_2(g) \rightarrow N_2(g) + 2H_2O(g) \][/tex]
2. The standard enthalpy change ([tex]\(\Delta H^{\circ}\)[/tex]) for this reaction is given as:
[tex]\[ \Delta H^{\circ} = -543.0 \ \frac{kJ}{mol} \][/tex]
This means that for every mole of [tex]\(N_2H_4\)[/tex] reacting with [tex]\(O_2\)[/tex] under standard conditions, 543.0 kJ of energy is released (hence the negative sign).
3. Since the given problem states that the reaction remains unchanged to connect with the goal reaction, the enthalpy change for the modified reaction is the same as the standard enthalpy change provided.
4. Therefore, the enthalpy for the modified reaction is:
[tex]\[ \Delta H^{\circ} = -543.0 \ \frac{kJ}{mol} \][/tex]
This concludes the detailed explanation. The final answer should include the sign and the magnitude of the enthalpy change with an appropriate unit and significant figures.
[tex]\[ \boxed{-543.0 \ \frac{kJ}{mol}} \][/tex]
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.