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Irina18 [472]
3 years ago
13

Instant cold packs, often used to ice athletic injuries on the field, contain ammonium nitrate and water separated by a thin pla

stic divider. When the divider is broken, the ammonium nitrate dissolves according to the following endothermic reaction: NH4NO3(s)---->NH+4(aq)+NO-3(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 degree C and the final temperature (after the solid dissolves) is 21.9 degree C.
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 degree C as the specific heat capacity.)

Express your answer to two significant figures and include the appropriate units.
Chemistry
1 answer:
RSB [31]3 years ago
4 0

<u>Answer:</u> The enthalpy change of the reaction is -27. kJ/mol

<u>Explanation:</u>

To calculate the mass of water, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of water = 1 g/mL

Volume of water = 25.0 mL

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{25.0mL}\\\\\text{Mass of water}=(1g/mL\times 25.0mL)=25g

To calculate the heat released by the reaction, we use the equation:

q=mc\Delta T

where,

q = heat released

m = Total mass = [1.25 + 25] = 26.25 g

c = heat capacity of water = 4.18 J/g°C

\Delta T = change in temperature = T_2-T_1=(21.9-25.8)^oC=-3.9^oC

Putting values in above equation, we get:

q=26.25g\tiimes 4.18J/g^oC\times (-3.9^oC)=-427.9J=-0.428kJ

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of ammonium nitrate = 1.25 g

Molar mass of ammonium nitrate = 80 g/mol

Putting values in above equation, we get:

\text{Moles of ammonium nitrate}=\frac{1.25g}{80g/mol}=0.0156mol

To calculate the enthalpy change of the reaction, we use the equation:

\Delta H_{rxn}=\frac{q}{n}

where,

q = amount of heat released = -0.428 kJ

n = number of moles = 0.0156 moles

\Delta H_{rxn} = enthalpy change of the reaction

Putting values in above equation, we get:

\Delta H_{rxn}=\frac{-0.428kJ}{0.0156mol}=-27.44kJ/mol

Hence, the enthalpy change of the reaction is -27. kJ/mol

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