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Ethyl chloride (C2H5Cl), is prepared by reaction of ethylene with hydrogen chloride :
C2H4(g) + HCl (g) → C2H5cl(g) ΔH = -72.3 kJ
What is the value of ΔE (in kJ), if 70 g of ethylene and 73 g of HCl are allowed to react at 300 K.
- a)-69.8
- b)-180.75
- c)-174.5
- d)-139.6
Correct answer is option 'D'. Can you explain this answer?
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Ethyl chloride (C2H5Cl), is prepared by reaction of ethylene with hydr...
2 mole HCl is consumed completely for 1 mole consumption of HCl
ΔU = ΔH – Δng RT
ΔU = ΔH – Δng RT
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Ethyl chloride (C2H5Cl), is prepared by reaction of ethylene with hydr...
To determine the value of E (in kJ) for the reaction of 70 g of ethylene (C2H4) and 73 g of HCl at 300 K, we need to calculate the heat released or absorbed during the reaction.
First, we need to calculate the moles of ethylene and HCl using their respective molar masses:
Molar mass of C2H4 = 2(12.01 g/mol) + 4(1.01 g/mol) = 28.05 g/mol
Moles of C2H4 = 70 g / 28.05 g/mol = 2.49 mol
Molar mass of HCl = 1.01 g/mol + 35.45 g/mol = 36.46 g/mol
Moles of HCl = 73 g / 36.46 g/mol = 2.00 mol
Since the reaction stoichiometry tells us that the ratio of C2H4 to HCl is 1:1, we can see that the limiting reactant is HCl, as we have exactly 2 moles of it.
Now we can use the given value of ΔH (heat change) for the reaction to calculate the value of E:
ΔH = -72.3 kJ (given)
Moles of HCl = 2.00 mol (from above)
ΔH = E / moles of HCl
Rearranging the equation, we can solve for E:
E = ΔH × moles of HCl = -72.3 kJ × 2.00 mol
E = -144.6 kJ
Therefore, the value of E is -144.6 kJ. However, the answer choices provided do not include this value. It is possible that there is an error in the answer choices or the given information.
First, we need to calculate the moles of ethylene and HCl using their respective molar masses:
Molar mass of C2H4 = 2(12.01 g/mol) + 4(1.01 g/mol) = 28.05 g/mol
Moles of C2H4 = 70 g / 28.05 g/mol = 2.49 mol
Molar mass of HCl = 1.01 g/mol + 35.45 g/mol = 36.46 g/mol
Moles of HCl = 73 g / 36.46 g/mol = 2.00 mol
Since the reaction stoichiometry tells us that the ratio of C2H4 to HCl is 1:1, we can see that the limiting reactant is HCl, as we have exactly 2 moles of it.
Now we can use the given value of ΔH (heat change) for the reaction to calculate the value of E:
ΔH = -72.3 kJ (given)
Moles of HCl = 2.00 mol (from above)
ΔH = E / moles of HCl
Rearranging the equation, we can solve for E:
E = ΔH × moles of HCl = -72.3 kJ × 2.00 mol
E = -144.6 kJ
Therefore, the value of E is -144.6 kJ. However, the answer choices provided do not include this value. It is possible that there is an error in the answer choices or the given information.
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Ethyl chloride (C2H5Cl), is prepared by reaction of ethylene with hydrogen chloride :C2H4(g) + HCl (g) → C2H5cl(g) ΔH = -72.3 kJWhat is the value of ΔE (in kJ), if 70 g of ethylene and 73 g of HCl are allowed to react at 300 K.a)-69.8b)-180.75c)-174.5d)-139.6Correct answer is option 'D'. Can you explain this answer?
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Ethyl chloride (C2H5Cl), is prepared by reaction of ethylene with hydrogen chloride :C2H4(g) + HCl (g) → C2H5cl(g) ΔH = -72.3 kJWhat is the value of ΔE (in kJ), if 70 g of ethylene and 73 g of HCl are allowed to react at 300 K.a)-69.8b)-180.75c)-174.5d)-139.6Correct answer is option 'D'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Ethyl chloride (C2H5Cl), is prepared by reaction of ethylene with hydrogen chloride :C2H4(g) + HCl (g) → C2H5cl(g) ΔH = -72.3 kJWhat is the value of ΔE (in kJ), if 70 g of ethylene and 73 g of HCl are allowed to react at 300 K.a)-69.8b)-180.75c)-174.5d)-139.6Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Ethyl chloride (C2H5Cl), is prepared by reaction of ethylene with hydrogen chloride :C2H4(g) + HCl (g) → C2H5cl(g) ΔH = -72.3 kJWhat is the value of ΔE (in kJ), if 70 g of ethylene and 73 g of HCl are allowed to react at 300 K.a)-69.8b)-180.75c)-174.5d)-139.6Correct answer is option 'D'. Can you explain this answer?.
Ethyl chloride (C2H5Cl), is prepared by reaction of ethylene with hydrogen chloride :C2H4(g) + HCl (g) → C2H5cl(g) ΔH = -72.3 kJWhat is the value of ΔE (in kJ), if 70 g of ethylene and 73 g of HCl are allowed to react at 300 K.a)-69.8b)-180.75c)-174.5d)-139.6Correct answer is option 'D'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Ethyl chloride (C2H5Cl), is prepared by reaction of ethylene with hydrogen chloride :C2H4(g) + HCl (g) → C2H5cl(g) ΔH = -72.3 kJWhat is the value of ΔE (in kJ), if 70 g of ethylene and 73 g of HCl are allowed to react at 300 K.a)-69.8b)-180.75c)-174.5d)-139.6Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Ethyl chloride (C2H5Cl), is prepared by reaction of ethylene with hydrogen chloride :C2H4(g) + HCl (g) → C2H5cl(g) ΔH = -72.3 kJWhat is the value of ΔE (in kJ), if 70 g of ethylene and 73 g of HCl are allowed to react at 300 K.a)-69.8b)-180.75c)-174.5d)-139.6Correct answer is option 'D'. Can you explain this answer?.
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