Question

Calculate the change in enthalpy for the reaction. (Use 1.0 g/mL as the density of the solution and 4.18 J/(g⋅∘C) as the specific heat capacity.) Express the enthalpy change in kilojoules per mole to two significant figures. ΔHrxn = nothing kJ/mol

Answer 1

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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)→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 ∘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.) Express your answer to two significant figures and include the appropriate units. ΔHrxn =   ??? kJ/mol

Answer:

26 kJ / mol

Explanation:

Given that;

The mass of NH₄NO₃ = 1.25 g

Number of moles of NH₄NO₃ = Mass of NH₄NO₃ / Molar Mass of NH₄NO₃

Number of moles of NH₄NO₃= 1.25 g / 80.043 g/mol

Number of moles of NH₄NO₃= 0.016 mol

Volume of solution = 25.0 mL

Density of Solution = 1.0g/mL

Since; $density = \frac {mass} {volume}$

Mass of Solution = Density x Volume

= 1.0g/mL × 25.0mL

= 25 g

Heat Generated (Q)  = $mc \delta T$

Q= 25g × 4.18 J/g°C x (25.8°C - 21.9°C)

Q =  407.55 J

Q = 407 × 10 ⁻³ kJ

Q = 0.40755 kJ

Δ $H_{rxn}$ = $\frac{Heat generated(Q)}{number of molesof NH_4NO_3}$

=  $\frac {0.40755 kJ}{ 0.016 mol}$

= 25.47 kJ/ mol

~ 26 kJ / mol

Therefore, the change in enthalpy for the reaction in kilojoules per mole = 26 kJ / mol