Would the change in enthalpy (ΔH) for the dissolution of urea in water be positive or negative?Homemade...
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Would the change in enthalpy (ΔH) for the dissolution of urea in water be positive or negative?
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$begingroup$
To test the properties of a fertilizer, 15.0 g of urea, NH2CONH2(s), is dissolved in 150mL of water in a simple calorimeter. A temperature change from 20.6 C to 17.8 C is measured. Calculate the molar enthalpy of solution for the fertilizer urea
I worked through this question by finding Q = mcΔT, and then dividing Q by the moles of urea present. I can tell the process is endothermic because ΔT is negative, however my answer for ΔH comes out as negative, which would only make sense if this was an exothermic reaction. I'm not sure where I am wrong to be honest.
Here is my work:
ΔH = (150 mL × 1g/mL × 4.18 J/gC × -2.8 C) ÷ (15 g ÷ 60.07 g) = -7030.59 J/mol, = -7.03 kJ/mol
TL;DR - question asks for ΔH of an endothermic process, not sure if my answer should be positive or negative
thermodynamics water aqueous-solution enthalpy
asked 5 hours ago
ZedEmZedEm
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$begingroup$
To test the properties of a fertilizer, 15.0 g of urea, NH2CONH2(s), is dissolved in 150mL of water in a simple calorimeter. A temperature change from 20.6 C to 17.8 C is measured. Calculate the molar enthalpy of solution for the fertilizer urea
I worked through this question by finding Q = mcΔT, and then dividing Q by the moles of urea present. I can tell the process is endothermic because ΔT is negative, however my answer for ΔH comes out as negative, which would only make sense if this was an exothermic reaction. I'm not sure where I am wrong to be honest.
Here is my work:
ΔH = (150 mL × 1g/mL × 4.18 J/gC × -2.8 C) ÷ (15 g ÷ 60.07 g) = -7030.59 J/mol, = -7.03 kJ/mol
TL;DR - question asks for ΔH of an endothermic process, not sure if my answer should be positive or negative
thermodynamics water aqueous-solution enthalpy
asked 5 hours ago
ZedEmZedEm
134
New contributor
ZedEm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
To test the properties of a fertilizer, 15.0 g of urea, NH2CONH2(s), is dissolved in 150mL of water in a simple calorimeter. A temperature change from 20.6 C to 17.8 C is measured. Calculate the molar enthalpy of solution for the fertilizer urea
I worked through this question by finding Q = mcΔT, and then dividing Q by the moles of urea present. I can tell the process is endothermic because ΔT is negative, however my answer for ΔH comes out as negative, which would only make sense if this was an exothermic reaction. I'm not sure where I am wrong to be honest.
Here is my work:
ΔH = (150 mL × 1g/mL × 4.18 J/gC × -2.8 C) ÷ (15 g ÷ 60.07 g) = -7030.59 J/mol, = -7.03 kJ/mol
TL;DR - question asks for ΔH of an endothermic process, not sure if my answer should be positive or negative
thermodynamics water aqueous-solution enthalpy
asked 5 hours ago
ZedEmZedEm
134
New contributor
ZedEm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
To test the properties of a fertilizer, 15.0 g of urea, NH2CONH2(s), is dissolved in 150mL of water in a simple calorimeter. A temperature change from 20.6 C to 17.8 C is measured. Calculate the molar enthalpy of solution for the fertilizer urea
I worked through this question by finding Q = mcΔT, and then dividing Q by the moles of urea present. I can tell the process is endothermic because ΔT is negative, however my answer for ΔH comes out as negative, which would only make sense if this was an exothermic reaction. I'm not sure where I am wrong to be honest.
Here is my work:
ΔH = (150 mL × 1g/mL × 4.18 J/gC × -2.8 C) ÷ (15 g ÷ 60.07 g) = -7030.59 J/mol, = -7.03 kJ/mol
TL;DR - question asks for ΔH of an endothermic process, not sure if my answer should be positive or negative
thermodynamics water aqueous-solution enthalpy
thermodynamics water aqueous-solution enthalpy
asked 5 hours ago
ZedEmZedEm
134
New contributor
ZedEm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
ZedEmZedEm
134
New contributor
ZedEm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
ZedEmZedEm
134
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ZedEm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
ZedEmZedEm
134
asked 5 hours ago
ZedEmZedEm
134
134
New contributor
ZedEm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
ZedEm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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$begingroup$
The sign of Q depends on the perspective. The water temperature decreased because it "lost" heat. The process of dissolving urea required energy, it "gained" energy. If I give you a penny, should that be +1 or -1 penny? Well, it depends who you ask.
In your answer, you are missing a negative sign in $Delta H=−Q$ the way you start out with $Q$ from the perspective of the water.
answered 4 hours ago
Karsten TheisKarsten Theis
4,989543
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$begingroup$
The sign of Q depends on the perspective. The water temperature decreased because it "lost" heat. The process of dissolving urea required energy, it "gained" energy. If I give you a penny, should that be +1 or -1 penny? Well, it depends who you ask.
In your answer, you are missing a negative sign in $Delta H=−Q$ the way you start out with $Q$ from the perspective of the water.
answered 4 hours ago
Karsten TheisKarsten Theis
4,989543
$endgroup$
add a comment |
$begingroup$
The sign of Q depends on the perspective. The water temperature decreased because it "lost" heat. The process of dissolving urea required energy, it "gained" energy. If I give you a penny, should that be +1 or -1 penny? Well, it depends who you ask.
In your answer, you are missing a negative sign in $Delta H=−Q$ the way you start out with $Q$ from the perspective of the water.
answered 4 hours ago
Karsten TheisKarsten Theis
4,989543
$endgroup$
add a comment |
$begingroup$
The sign of Q depends on the perspective. The water temperature decreased because it "lost" heat. The process of dissolving urea required energy, it "gained" energy. If I give you a penny, should that be +1 or -1 penny? Well, it depends who you ask.
In your answer, you are missing a negative sign in $Delta H=−Q$ the way you start out with $Q$ from the perspective of the water.
answered 4 hours ago
Karsten TheisKarsten Theis
4,989543
$endgroup$
The sign of Q depends on the perspective. The water temperature decreased because it "lost" heat. The process of dissolving urea required energy, it "gained" energy. If I give you a penny, should that be +1 or -1 penny? Well, it depends who you ask.
In your answer, you are missing a negative sign in $Delta H=−Q$ the way you start out with $Q$ from the perspective of the water.
answered 4 hours ago
Karsten TheisKarsten Theis
4,989543
answered 4 hours ago
Karsten TheisKarsten Theis
4,989543
answered 4 hours ago
Karsten TheisKarsten Theis
4,989543
answered 4 hours ago
Karsten TheisKarsten Theis
4,989543
4,989543
add a comment |
add a comment |
ZedEm is a new contributor. Be nice, and check out our Code of Conduct.
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