Question

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 degrees C and the final temperature (after the solid dissolves) is 21.9 degrees C. Calculate the change in enthalpy for the reaction. (Use 1.0g/mL as the density of the solution and 4.18 J/g . degrees C as the specific heat capacity.) Express the answer to TWO significant figures.

Answer 1

Answer:

The change in enthalpy for the reaction is 27 kJ/mol.

Explanation:

Volume of the solution = v = 25.0 mL

Mass of the solution = m

Density of the solution = d = 1.0 g/mL

$mass= density\times volume$

$m=d\timees v=1.0 g/mL\times 25.0 mL=25 g$

Heat capacity of the solution = C = 4.18 J/g°C

The initial temperature of solution,$T_1$ = 25.8°C

The final temperature of solution,$T_2$  = 21.8°C

Heat loss of the solution = Q

$Q=mC\Delta T=mc(T_2-T_1)$

$Q=25.0 g\times 4.18 J/g^oC\times ( 21.8°C- 25.8°C)$

$Q=-418 J$

Negative sign indicates that energy is released by the solution.

Energy gained by the ammonium nitrate  = Q'

Q'= -Q = 418 J (Conservation of energy)

Mass of ammonium nitrate = 1.25 g

Moles of ammonium nitrate ,n= $\frac{1.25 g}{80 g/mol}=0.015625 mol$

Enthalpy change of reaction = $\frac{Q'}{n}$

$=\frac{418 J}{0.015625 mol}=26752 J/mol=26.752 kJ/mol\approx 27 kJ/mol$