Question

A 50.0 mL of 0.88 M H2O2 and 10.0 mL of 0.50 M Fe(NO3)3 were combined and a temperature change of 7.47 was observed. The specific heat of water is 4.18 J/(g * ∘C) Calculate the heat of reaction (in kilojoules). Record your answer with the proper significant figures and include the correct sign if needed. Assume the density and specific heat of the solution are the same as that of water. kJ

Answer 1

Answer:

The heat of reaction -1.9 kilo Joules.

Explanation:

Volume of the hydrogen peroxide solution = 50.0 mL

Volume of the ferric nitrate solution = 10.0 mL

Total volume of the solution = 50.0 mL + 10.0 mL = 60.0 mL

Mass of the final solution = m

Density of water = density of the final solution = d = 1 g/mL

$Mass=density\times Volume$

$m=1 g/ml\times 60.0 ml=60.0 g$

Heat capacity of the mixture = c = 4.18 J/g°C

Change in temperature of the mixture = ΔT = 7.47°C

Heat absorbed by the mixture = Q

$Q=mc\Delta T$

$Q=60.0 g\times 4.18 J/g ^oC\times 7.47 ^oC=1873.48 J=1.873 kJ\approx 1.9 kJ$

Heat release due to reaction = -Q'

-Q' =Q (law of conservation of enrage)

Q'= -Q = -1.9 kJ

The heat of reaction -1.9 kilo Joules.

Answer 2

The heat of reaction is mathematically given as

Q'= -Q = -1.873 kJ

What is the heat of reaction?

Question Parameter(s):

Generally, the equation for the mass   is mathematically given as

Mass=density*Volume

Therefore

m=1 *60.0 ml

m=60.0 g

Where

Q=mcdT

Q=60.0 *4.18 * 7.47

Q=1873.48 J

Q=1.873 kJ

In conclusion, Heat released is

Q'= -Q = -1.873 kJ

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