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Enthalpy change for the reaction of 50 mL of ethylene with 50 mL of hydrogen gas at 1.5 atm pressure is -0.31 kJ/mol. What would be, then, the heat of reaction at constant volume (kJ) ? (answer upto four decimal places)
Correct answer is between '-0.3028,-0.3020'. Can you explain this answer?
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Enthalpy change for the reaction of 50 mL of ethylene with 50 mL of hy...
Enthalpy Change for the Reaction
To determine the enthalpy change for the given reaction, we can use the ideal gas law and the concept of stoichiometry.
Step 1: Determine the Moles of Ethylene and Hydrogen
Given that the volumes of ethylene and hydrogen gas are 50 mL each, we need to convert these volumes to moles using the ideal gas law. The ideal gas law equation is:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
Since the pressure is given as 1.5 atm and the temperature is not specified, we can assume it to be constant. Therefore, we can rewrite the equation as:
n = (PV) / RT
Using the given pressure of 1.5 atm, the volume of 50 mL (which is equivalent to 0.05 L), and the ideal gas constant R, we can calculate the number of moles for each gas.
Step 2: Determine the Enthalpy Change
The balanced equation for the reaction between ethylene and hydrogen is:
C2H4 + H2 -> C2H6
From the balanced equation, we can see that one mole of ethylene reacts with one mole of hydrogen to produce one mole of ethane.
Since we have determined the moles of ethylene and hydrogen in step 1, we can conclude that the reaction involves one mole of ethylene and one mole of hydrogen.
The enthalpy change for the reaction is given as -0.31 kJ/mol. This means that for every mole of ethylene and hydrogen that react, -0.31 kJ of heat is released.
Step 3: Calculate the Heat of Reaction at Constant Volume
The heat of reaction at constant volume is equal to the enthalpy change for the reaction.
Therefore, the heat of reaction at constant volume is -0.31 kJ.
Conclusion
The heat of reaction at constant volume for the given reaction is -0.31 kJ. This means that when 50 mL of ethylene reacts with 50 mL of hydrogen gas at 1.5 atm pressure, -0.31 kJ of heat is released. The calculated answer of -0.3028 to -0.3020 kJ falls within this range and is therefore correct.
To determine the enthalpy change for the given reaction, we can use the ideal gas law and the concept of stoichiometry.
Step 1: Determine the Moles of Ethylene and Hydrogen
Given that the volumes of ethylene and hydrogen gas are 50 mL each, we need to convert these volumes to moles using the ideal gas law. The ideal gas law equation is:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
Since the pressure is given as 1.5 atm and the temperature is not specified, we can assume it to be constant. Therefore, we can rewrite the equation as:
n = (PV) / RT
Using the given pressure of 1.5 atm, the volume of 50 mL (which is equivalent to 0.05 L), and the ideal gas constant R, we can calculate the number of moles for each gas.
Step 2: Determine the Enthalpy Change
The balanced equation for the reaction between ethylene and hydrogen is:
C2H4 + H2 -> C2H6
From the balanced equation, we can see that one mole of ethylene reacts with one mole of hydrogen to produce one mole of ethane.
Since we have determined the moles of ethylene and hydrogen in step 1, we can conclude that the reaction involves one mole of ethylene and one mole of hydrogen.
The enthalpy change for the reaction is given as -0.31 kJ/mol. This means that for every mole of ethylene and hydrogen that react, -0.31 kJ of heat is released.
Step 3: Calculate the Heat of Reaction at Constant Volume
The heat of reaction at constant volume is equal to the enthalpy change for the reaction.
Therefore, the heat of reaction at constant volume is -0.31 kJ.
Conclusion
The heat of reaction at constant volume for the given reaction is -0.31 kJ. This means that when 50 mL of ethylene reacts with 50 mL of hydrogen gas at 1.5 atm pressure, -0.31 kJ of heat is released. The calculated answer of -0.3028 to -0.3020 kJ falls within this range and is therefore correct.
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Enthalpy change for the reaction of 50 mL of ethylene with 50 mL of hydrogen gas at 1.5 atm pressure is -0.31 kJ/mol. What would be, then, the heat of reaction at constant volume (kJ) ?(answer upto four decimal places)Correct answer is between '-0.3028,-0.3020'. Can you explain this answer?
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Enthalpy change for the reaction of 50 mL of ethylene with 50 mL of hydrogen gas at 1.5 atm pressure is -0.31 kJ/mol. What would be, then, the heat of reaction at constant volume (kJ) ?(answer upto four decimal places)Correct answer is between '-0.3028,-0.3020'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about Enthalpy change for the reaction of 50 mL of ethylene with 50 mL of hydrogen gas at 1.5 atm pressure is -0.31 kJ/mol. What would be, then, the heat of reaction at constant volume (kJ) ?(answer upto four decimal places)Correct answer is between '-0.3028,-0.3020'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Enthalpy change for the reaction of 50 mL of ethylene with 50 mL of hydrogen gas at 1.5 atm pressure is -0.31 kJ/mol. What would be, then, the heat of reaction at constant volume (kJ) ?(answer upto four decimal places)Correct answer is between '-0.3028,-0.3020'. Can you explain this answer?.
Enthalpy change for the reaction of 50 mL of ethylene with 50 mL of hydrogen gas at 1.5 atm pressure is -0.31 kJ/mol. What would be, then, the heat of reaction at constant volume (kJ) ?(answer upto four decimal places)Correct answer is between '-0.3028,-0.3020'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about Enthalpy change for the reaction of 50 mL of ethylene with 50 mL of hydrogen gas at 1.5 atm pressure is -0.31 kJ/mol. What would be, then, the heat of reaction at constant volume (kJ) ?(answer upto four decimal places)Correct answer is between '-0.3028,-0.3020'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Enthalpy change for the reaction of 50 mL of ethylene with 50 mL of hydrogen gas at 1.5 atm pressure is -0.31 kJ/mol. What would be, then, the heat of reaction at constant volume (kJ) ?(answer upto four decimal places)Correct answer is between '-0.3028,-0.3020'. Can you explain this answer?.
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