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Would the change in enthalpy (ΔH) for the dissolution of urea in water be positive or negative?
Homemade reactor for water coolingHow do I calculate the enthalpy change when a gas is being used to heat water?Calculating amount of ice required for heat lossHow to calculate the heat of dissolution from a calorimeter experiment?Heat given off from an electrochemical cell compared to mixing reactantsHow to determine whether the enthalpy of solution is positive or negative by calorimetry?Calculating heat of combustion via calorimetryCalculating enthalpy of dissolutionheat of fusion in an equationCalculating the heat of reaction for Mg metal and HCl
$begingroup$
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
New contributor
$endgroup$
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
New contributor
$endgroup$
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
New contributor
$endgroup$
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
thermodynamics water aqueous-solution enthalpy
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edited 1 hour ago
Loong♦
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asked 8 hours ago
ZedEmZedEm
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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.
$endgroup$
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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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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 7 hours ago
Karsten TheisKarsten Theis
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ZedEm is a new contributor. Be nice, and check out our Code of Conduct.
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