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4 March, 22:26

Instant cold packs, often used to ice athletic injuries on the field, contain ammonium nitrate and water separated by a thin plastic divider. When the divider is broken, the ammonium nitrate dissolves according to the following endothermic reaction: NH4NO3 (s) →NH4 + (aq) + NO3 - (aq) In order to measure the enthalpy change for this reaction, 1.25 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 21.9 ∘C. Part A Calculate the change in enthalpy for the reaction in kilojoules per mole. (Use 1.0g/mL as the density of the solution and 4.18J/g⋅∘C as the specific heat capacity.)

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  1. 5 March, 01:02
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    26.08 kJ

    Explanation:

    An endothermic reaction occurs when the heat is absorbed by the system, it means that it's a positive heat, and consequently, the enthalpy change will be positive too. Because that isn't happening a phase change, the heat transferred is sensitive, and it can be calculated by:

    Q = m*c*ΔT

    Where Q is the heat, m is the mass of the water (because the change is happening in it), c is the specific heat of water, and ΔT is the temperature change (final - initial).

    m = volume*density

    m = 25.00 mL * 1 g/mL

    m = 25 g

    The water is losing heat, so for it, the heat must be negative, because it will be an exothermic process. The heat for the reaction will be the same that is lost by water because the energy must be conserved (first law of thermodynamics):

    Q = 25*4.18 * (21.9 - 25.8)

    Q = - 407.55 J

    Q = + 407.55 J (for the endothermic reaction given)

    The enthalpy change is the heat divided by the number of moles of the substance that is dissolved. The molar mass of NH₄NO₃ is:

    2*14 g/mol of N + 4*1 g/mol of H + 3*16 g/mol of O = 80 g/mol

    The number of moles is the mass divided by the molar mass:

    n = 1.25/80

    n = 0.015625

    So, the change in enthalpy is:

    ΔH = + 407.55/0.015625

    ΔH = 26,083.2 J

    ΔH = 26.08 kJ
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