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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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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:



work for enthalpy problem




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










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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:



    work for enthalpy problem




    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










    share|improve this question







    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.







    $endgroup$















      2












      2








      2





      $begingroup$



      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:



      work for enthalpy problem




      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










      share|improve this question







      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:



      work for enthalpy problem




      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






      share|improve this question







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      ZedEm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      share|improve this question






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      asked 4 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.






          share|improve this answer









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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.






            share|improve this answer









            $endgroup$


















              2












              $begingroup$

              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









              $endgroup$
















                2












                2








                2





                $begingroup$

                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









                $endgroup$



                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.







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 3 hours ago









                Karsten TheisKarsten Theis

                4,964543




                4,964543






















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