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For a gaseous reaction, A(g)+3B(g)?3C(g)+3D(g), ?deltaE is 17 kcal at 270C. Assuming R=2 cal K-1 mol-, the value of ?deltaH for the above reaction?
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For a gaseous reaction, A(g)+3B(g)?3C(g)+3D(g), ?deltaE is 17 kcal at ...
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For a gaseous reaction, A(g)+3B(g)?3C(g)+3D(g), ?deltaE is 17 kcal at ...
Explanation:
The enthalpy change (ΔH) for a reaction can be calculated using the equation:
ΔH = ΔE + ΔnRT
Where:
ΔH = change in enthalpy
ΔE = change in internal energy
Δn = change in the number of moles of gas
R = ideal gas constant
T = temperature in Kelvin
Given:
ΔE = 17 kcal
R = 2 cal K-1 mol-1
T = 270°C = 543 K
We need to determine the value of ΔH for the reaction A(g) + 3B(g) → 3C(g) + 3D(g).
Step 1: Calculate Δn
Δn = (sum of moles of products) - (sum of moles of reactants)
= (3+3) - (1+3)
= 2 moles
Step 2: Convert ΔE to cal
1 kcal = 1000 cal
ΔE = 17 kcal = 17,000 cal
Step 3: Calculate ΔH
ΔH = ΔE + ΔnRT
= 17,000 cal + (2 mol)(2 cal K-1 mol-1)(543 K)
= 17,000 cal + 2,172 cal
= 19,172 cal
Therefore, the value of ΔH for the reaction is 19,172 cal.
Summary:
- The enthalpy change (ΔH) for a reaction can be calculated using the equation ΔH = ΔE + ΔnRT.
- Δn is the change in the number of moles of gas.
- Given the values of ΔE, R, and T, we can calculate ΔH.
- In this case, ΔE is given as 17 kcal, R is 2 cal K-1 mol-1, and T is 270°C (543 K).
- We calculate Δn to be 2 moles.
- Converting ΔE to cal, we get 17,000 cal.
- Plugging the values into the equation, we find ΔH to be 19,172 cal.
The enthalpy change (ΔH) for a reaction can be calculated using the equation:
ΔH = ΔE + ΔnRT
Where:
ΔH = change in enthalpy
ΔE = change in internal energy
Δn = change in the number of moles of gas
R = ideal gas constant
T = temperature in Kelvin
Given:
ΔE = 17 kcal
R = 2 cal K-1 mol-1
T = 270°C = 543 K
We need to determine the value of ΔH for the reaction A(g) + 3B(g) → 3C(g) + 3D(g).
Step 1: Calculate Δn
Δn = (sum of moles of products) - (sum of moles of reactants)
= (3+3) - (1+3)
= 2 moles
Step 2: Convert ΔE to cal
1 kcal = 1000 cal
ΔE = 17 kcal = 17,000 cal
Step 3: Calculate ΔH
ΔH = ΔE + ΔnRT
= 17,000 cal + (2 mol)(2 cal K-1 mol-1)(543 K)
= 17,000 cal + 2,172 cal
= 19,172 cal
Therefore, the value of ΔH for the reaction is 19,172 cal.
Summary:
- The enthalpy change (ΔH) for a reaction can be calculated using the equation ΔH = ΔE + ΔnRT.
- Δn is the change in the number of moles of gas.
- Given the values of ΔE, R, and T, we can calculate ΔH.
- In this case, ΔE is given as 17 kcal, R is 2 cal K-1 mol-1, and T is 270°C (543 K).
- We calculate Δn to be 2 moles.
- Converting ΔE to cal, we get 17,000 cal.
- Plugging the values into the equation, we find ΔH to be 19,172 cal.
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For a gaseous reaction, A(g)+3B(g)?3C(g)+3D(g), ?deltaE is 17 kcal at 270C. Assuming R=2 cal K-1 mol-, the value of ?deltaH for the above reaction?
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For a gaseous reaction, A(g)+3B(g)?3C(g)+3D(g), ?deltaE is 17 kcal at 270C. Assuming R=2 cal K-1 mol-, the value of ?deltaH for the above reaction? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about For a gaseous reaction, A(g)+3B(g)?3C(g)+3D(g), ?deltaE is 17 kcal at 270C. Assuming R=2 cal K-1 mol-, the value of ?deltaH for the above reaction? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For a gaseous reaction, A(g)+3B(g)?3C(g)+3D(g), ?deltaE is 17 kcal at 270C. Assuming R=2 cal K-1 mol-, the value of ?deltaH for the above reaction?.
For a gaseous reaction, A(g)+3B(g)?3C(g)+3D(g), ?deltaE is 17 kcal at 270C. Assuming R=2 cal K-1 mol-, the value of ?deltaH for the above reaction? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about For a gaseous reaction, A(g)+3B(g)?3C(g)+3D(g), ?deltaE is 17 kcal at 270C. Assuming R=2 cal K-1 mol-, the value of ?deltaH for the above reaction? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For a gaseous reaction, A(g)+3B(g)?3C(g)+3D(g), ?deltaE is 17 kcal at 270C. Assuming R=2 cal K-1 mol-, the value of ?deltaH for the above reaction?.
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