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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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4

$begingroup$

To test the properties of a fertilizer, $15.0 mathrm g$ of urea, $ceNH2CONH2(s)$, is dissolved in $150 mathrmmL$ of water in a simple calorimeter. A temperature change from $20.6 mathrm^circ C$ to $17.8 mathrm^circ C$ is measured. Calculate the molar enthalpy of solution for the fertilizer urea

I worked through this question by finding $Q = mcDelta T$, and then dividing $Q$ by the moles of urea present. I can tell the process is endothermic because $Delta T$ is negative, however my answer for $Delta H$ 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:

$$Delta H = (150 mathrmmL times 1 mathrmg/mL times 4.18 mathrmJ/(g ^circ C) times -2.8 mathrm^circ C) / (15 mathrm g / 60.07 mathrm g) = -7030.59 mathrmJ/mol = -7.03 mathrmkJ/mol$$

TL;DR - question asks for $Delta H$ of an endothermic process, not sure if my answer should be positive or negative

thermodynamics water aqueous-solution enthalpy

share|improve this question

edited Apr 26 at 5:34

Loong♦

34.6k886184

asked Apr 25 at 22:21

ZedEmZedEm

234

$endgroup$

add a comment | 

4

$begingroup$

To test the properties of a fertilizer, $15.0 mathrm g$ of urea, $ceNH2CONH2(s)$, is dissolved in $150 mathrmmL$ of water in a simple calorimeter. A temperature change from $20.6 mathrm^circ C$ to $17.8 mathrm^circ C$ is measured. Calculate the molar enthalpy of solution for the fertilizer urea

I worked through this question by finding $Q = mcDelta T$, and then dividing $Q$ by the moles of urea present. I can tell the process is endothermic because $Delta T$ is negative, however my answer for $Delta H$ 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:

$$Delta H = (150 mathrmmL times 1 mathrmg/mL times 4.18 mathrmJ/(g ^circ C) times -2.8 mathrm^circ C) / (15 mathrm g / 60.07 mathrm g) = -7030.59 mathrmJ/mol = -7.03 mathrmkJ/mol$$

TL;DR - question asks for $Delta H$ of an endothermic process, not sure if my answer should be positive or negative

thermodynamics water aqueous-solution enthalpy

share|improve this question

edited Apr 26 at 5:34

Loong♦

34.6k886184

asked Apr 25 at 22:21

ZedEmZedEm

234

$endgroup$

add a comment | 

$begingroup$

To test the properties of a fertilizer, $15.0 mathrm g$ of urea, $ceNH2CONH2(s)$, is dissolved in $150 mathrmmL$ of water in a simple calorimeter. A temperature change from $20.6 mathrm^circ C$ to $17.8 mathrm^circ C$ is measured. Calculate the molar enthalpy of solution for the fertilizer urea

I worked through this question by finding $Q = mcDelta T$, and then dividing $Q$ by the moles of urea present. I can tell the process is endothermic because $Delta T$ is negative, however my answer for $Delta H$ 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:

$$Delta H = (150 mathrmmL times 1 mathrmg/mL times 4.18 mathrmJ/(g ^circ C) times -2.8 mathrm^circ C) / (15 mathrm g / 60.07 mathrm g) = -7030.59 mathrmJ/mol = -7.03 mathrmkJ/mol$$

TL;DR - question asks for $Delta H$ of an endothermic process, not sure if my answer should be positive or negative

thermodynamics water aqueous-solution enthalpy

share|improve this question

edited Apr 26 at 5:34

Loong♦

34.6k886184

asked Apr 25 at 22:21

ZedEmZedEm

234

$endgroup$

share|improve this question

edited Apr 26 at 5:34

Loong♦

34.6k886184

asked Apr 25 at 22:21

ZedEmZedEm

234

share|improve this question

edited Apr 26 at 5:34

Loong♦

34.6k886184

asked Apr 25 at 22:21

ZedEmZedEm

234

edited Apr 26 at 5:34

Loong♦

34.6k886184

edited Apr 26 at 5:34

Loong♦

34.6k886184

asked Apr 25 at 22:21

ZedEmZedEm

234

asked Apr 25 at 22:21

ZedEmZedEm

234

1 Answer
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active

oldest

votes

4

$begingroup$

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.

share|improve this answer

answered Apr 25 at 23:02

Karsten TheisKarsten Theis

5,621745

$endgroup$

add a comment | 

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4

$begingroup$

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.

share|improve this answer

answered Apr 25 at 23:02

Karsten TheisKarsten Theis

5,621745

$endgroup$

add a comment | 

4

$begingroup$

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.

share|improve this answer

answered Apr 25 at 23:02

Karsten TheisKarsten Theis

5,621745

$endgroup$

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.

share|improve this answer

answered Apr 25 at 23:02

Karsten TheisKarsten Theis

5,621745

$endgroup$

share|improve this answer

answered Apr 25 at 23:02

Karsten TheisKarsten Theis

5,621745

answered Apr 25 at 23:02

Karsten TheisKarsten Theis

5,621745

answered Apr 25 at 23:02

Karsten TheisKarsten Theis

5,621745