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NH4NO3(s) NH4+(aq) + NO3−(aq)

In order to measure the enthalpy change for this reaction above, 1.07 g of NH4NO3 is dissolved in enough water to make 25.0 mL of solution. The initial temperature is 25.8°C and the final temperature (after the solid dissolves) is 22.4°C. Calculate the change in enthalpy for the reaction in kJ. (Use 1.0 g/mL as the density of the solution and 4.18 J/g · °C as the specific heat capacity.)

We've been doing a lot of energy problems, but I'm not sure how to begin this one. I don't want the answer, just the set up so I know what I'm doing. Thanks. :)

  • chemistry - ,

    The rise is T for H2O is your route to heat involved.
    q = mass H2O x specific heat H2O x delta T.
    Then q/1.07 = delta H/gram.
    Usually these are quoted in kJ/mol.
    (delta H/gram NH4NO3) x molar mass NH4NO3 x (1 kJ/1000 J) = ?
    Note: Since T went down you know it was cooled and means an endothermic reaction so the sign of delta H is +.

  • chemistry - ,

    Would the answer be 27kJ?

  • chemistry - ,

    I think you can report more than 2 significant figures.
    I calculated 26.6 kJ/mol.

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