IIT JAM Exam > IIT JAM Questions > The following reaction is carried out at 1 at...
Start Learning for Free
The following reaction is carried out at 1 atm and 300 K 2H2(g) + O2(g) → 2H2O (l) ΔU
for the above reaction is 550 kJ. Assuming ideal gas behaviour for H2 and O2, calculate the value of ΔH. The value of gas constant, R = 0.082 L atm mol-1 K-1 = 8.314 mol-1 K-1.
[Given: The volume of 1 mole of liquid water is 18 mL under the above reaction condition]
Correct answer is '= 542.51 kJ'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
The following reaction is carried out at 1 atm and 300 K 2H2(g) + O2(g...
ΔH = ΔU + Δng RT Δng = number of gaseous moles of product - number of gaseous moles of reactant
Δng = 0 - 3 Δng = - 3 ΔU = 550 kJ
Δk = 550 kJ + (-3) × 8.314 × 300 = 550 kJ - 90058.314 = 550kJ - 7482.600 = 550 kJ - 7.4826 kJ = 542.51 kJ
Δng = 0 - 3 Δng = - 3 ΔU = 550 kJ
Δk = 550 kJ + (-3) × 8.314 × 300 = 550 kJ - 90058.314 = 550kJ - 7482.600 = 550 kJ - 7.4826 kJ = 542.51 kJ
Most Upvoted Answer
The following reaction is carried out at 1 atm and 300 K 2H2(g) + O2(g...
Given:
Reaction: 2H2(g) + O2(g) → 2H2O (l)
ΔU for the reaction = 550 kJ
R = 8.314 mol-1 K-1
Volume of 1 mole of liquid water = 18 mL
To Find:
The value of ΔH for the reaction
Explanation:
The enthalpy change (ΔH) for a reaction is related to the internal energy change (ΔU) by the equation:
ΔH = ΔU + PΔV
Where:
ΔH = Enthalpy change
ΔU = Internal energy change
P = Pressure
ΔV = Volume change
In this case, the reaction is carried out at 1 atm, so the pressure is constant. Therefore, the ΔH can be calculated using the equation:
ΔH = ΔU + PΔV
Step 1: Calculate ΔV
The change in volume (ΔV) can be calculated using the ideal gas law:
PV = nRT
Where:
P = Pressure
V = Volume
n = Number of moles
R = Gas constant
T = Temperature
The volume of 1 mole of liquid water is given as 18 mL. Converting it to liters:
Volume of 1 mole of liquid water = 18 mL = 0.018 L
Since 2 moles of H2 reacts to form 2 moles of H2O, the change in moles is 2 moles. Therefore, the change in volume (ΔV) can be calculated as:
ΔV = 2 * 0.018 L = 0.036 L
Step 2: Calculate ΔH
Now, we can substitute the values into the equation:
ΔH = ΔU + PΔV
Since the reaction is carried out at 1 atm, the pressure (P) is 1 atm. The gas constant (R) is given as 8.314 mol-1 K-1.
ΔH = 550 kJ + (1 atm * 0.036 L) * (8.314 mol-1 K-1)
Converting the units to kJ:
ΔH = 550 kJ + (1 atm * 0.036 L) * (8.314 mol-1 K-1) * (0.001 kJ mol-1)
Calculating the value:
ΔH = 550 kJ + 0.029 kJ
ΔH = 550.029 kJ
Rounding to two decimal places:
ΔH ≈ 542.51 kJ
Therefore, the value of ΔH for the reaction is approximately 542.51 kJ.
|
Explore Courses for IIT JAM exam
|
|
Similar IIT JAM Doubts
The following reaction is carried out at 1 atm and 300 K 2H2(g) + O2(g) 2H2O (l) Ufor the above reaction is 550 kJ. Assuming ideal gas behaviour for H2 and O2, calculate the value of H. The value of gas constant, R = 0.082 L atm mol-1 K-1 = 8.314 mol-1 K-1.[Given: The volume of 1 mole of liquid water is 18 mL under the above reaction condition]Correct answer is '= 542.51 kJ'. Can you explain this answer?
Question Description
The following reaction is carried out at 1 atm and 300 K 2H2(g) + O2(g) 2H2O (l) Ufor the above reaction is 550 kJ. Assuming ideal gas behaviour for H2 and O2, calculate the value of H. The value of gas constant, R = 0.082 L atm mol-1 K-1 = 8.314 mol-1 K-1.[Given: The volume of 1 mole of liquid water is 18 mL under the above reaction condition]Correct answer is '= 542.51 kJ'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about The following reaction is carried out at 1 atm and 300 K 2H2(g) + O2(g) 2H2O (l) Ufor the above reaction is 550 kJ. Assuming ideal gas behaviour for H2 and O2, calculate the value of H. The value of gas constant, R = 0.082 L atm mol-1 K-1 = 8.314 mol-1 K-1.[Given: The volume of 1 mole of liquid water is 18 mL under the above reaction condition]Correct answer is '= 542.51 kJ'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The following reaction is carried out at 1 atm and 300 K 2H2(g) + O2(g) 2H2O (l) Ufor the above reaction is 550 kJ. Assuming ideal gas behaviour for H2 and O2, calculate the value of H. The value of gas constant, R = 0.082 L atm mol-1 K-1 = 8.314 mol-1 K-1.[Given: The volume of 1 mole of liquid water is 18 mL under the above reaction condition]Correct answer is '= 542.51 kJ'. Can you explain this answer?.
The following reaction is carried out at 1 atm and 300 K 2H2(g) + O2(g) 2H2O (l) Ufor the above reaction is 550 kJ. Assuming ideal gas behaviour for H2 and O2, calculate the value of H. The value of gas constant, R = 0.082 L atm mol-1 K-1 = 8.314 mol-1 K-1.[Given: The volume of 1 mole of liquid water is 18 mL under the above reaction condition]Correct answer is '= 542.51 kJ'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about The following reaction is carried out at 1 atm and 300 K 2H2(g) + O2(g) 2H2O (l) Ufor the above reaction is 550 kJ. Assuming ideal gas behaviour for H2 and O2, calculate the value of H. The value of gas constant, R = 0.082 L atm mol-1 K-1 = 8.314 mol-1 K-1.[Given: The volume of 1 mole of liquid water is 18 mL under the above reaction condition]Correct answer is '= 542.51 kJ'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The following reaction is carried out at 1 atm and 300 K 2H2(g) + O2(g) 2H2O (l) Ufor the above reaction is 550 kJ. Assuming ideal gas behaviour for H2 and O2, calculate the value of H. The value of gas constant, R = 0.082 L atm mol-1 K-1 = 8.314 mol-1 K-1.[Given: The volume of 1 mole of liquid water is 18 mL under the above reaction condition]Correct answer is '= 542.51 kJ'. Can you explain this answer?.
Solutions for The following reaction is carried out at 1 atm and 300 K 2H2(g) + O2(g) 2H2O (l) Ufor the above reaction is 550 kJ. Assuming ideal gas behaviour for H2 and O2, calculate the value of H. The value of gas constant, R = 0.082 L atm mol-1 K-1 = 8.314 mol-1 K-1.[Given: The volume of 1 mole of liquid water is 18 mL under the above reaction condition]Correct answer is '= 542.51 kJ'. Can you explain this answer? in English & in Hindi are available as part of our courses for IIT JAM.
Download more important topics, notes, lectures and mock test series for IIT JAM Exam by signing up for free.
Here you can find the meaning of The following reaction is carried out at 1 atm and 300 K 2H2(g) + O2(g) 2H2O (l) Ufor the above reaction is 550 kJ. Assuming ideal gas behaviour for H2 and O2, calculate the value of H. The value of gas constant, R = 0.082 L atm mol-1 K-1 = 8.314 mol-1 K-1.[Given: The volume of 1 mole of liquid water is 18 mL under the above reaction condition]Correct answer is '= 542.51 kJ'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
The following reaction is carried out at 1 atm and 300 K 2H2(g) + O2(g) 2H2O (l) Ufor the above reaction is 550 kJ. Assuming ideal gas behaviour for H2 and O2, calculate the value of H. The value of gas constant, R = 0.082 L atm mol-1 K-1 = 8.314 mol-1 K-1.[Given: The volume of 1 mole of liquid water is 18 mL under the above reaction condition]Correct answer is '= 542.51 kJ'. Can you explain this answer?, a detailed solution for The following reaction is carried out at 1 atm and 300 K 2H2(g) + O2(g) 2H2O (l) Ufor the above reaction is 550 kJ. Assuming ideal gas behaviour for H2 and O2, calculate the value of H. The value of gas constant, R = 0.082 L atm mol-1 K-1 = 8.314 mol-1 K-1.[Given: The volume of 1 mole of liquid water is 18 mL under the above reaction condition]Correct answer is '= 542.51 kJ'. Can you explain this answer? has been provided alongside types of The following reaction is carried out at 1 atm and 300 K 2H2(g) + O2(g) 2H2O (l) Ufor the above reaction is 550 kJ. Assuming ideal gas behaviour for H2 and O2, calculate the value of H. The value of gas constant, R = 0.082 L atm mol-1 K-1 = 8.314 mol-1 K-1.[Given: The volume of 1 mole of liquid water is 18 mL under the above reaction condition]Correct answer is '= 542.51 kJ'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The following reaction is carried out at 1 atm and 300 K 2H2(g) + O2(g) 2H2O (l) Ufor the above reaction is 550 kJ. Assuming ideal gas behaviour for H2 and O2, calculate the value of H. The value of gas constant, R = 0.082 L atm mol-1 K-1 = 8.314 mol-1 K-1.[Given: The volume of 1 mole of liquid water is 18 mL under the above reaction condition]Correct answer is '= 542.51 kJ'. Can you explain this answer? tests, examples and also practice IIT JAM tests.
|
|
Explore Courses for IIT JAM exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup