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Help with chemistry please - heat capacity?
The heat evolved in the decomposition of 7.647 g of ammonium nitrate can be
measured in a bomb calorimeter. The temperature of the calorimeter, which contains
415 g of water, increases from 18.90 °C to 20.72 °C. The heat capacity of the bomb
calorimeter is 155 J/K. What is ΔH for the reaction, assuming that the specific heat
capacity of the pure water is 4.18 J/g/°C.
2 Answers
- 9 years agoFavorite Answer
change in temperature = delta T = 20.72-18.9 = 1.82
q(water) = m(water) X c(water) X delta T = 415 X 4.18 X 1.82 = 3157.154 j
q(bomb) = c (bomb) X delta T = 155 X 1.82 = 282.1 j
q(decomposition) = -(3157.154 + 282.1) = -3439.254 j
molecular mass of ammonium nitrate = NH4NO3 = 80 g/mol
no. of moles = 7.647/80 = 0.095
so q(decomposition) per mole = -3439.254/0.095 = -36202.674 j/mole
- 9 years ago
q = mc^T
where:
q = energy lost (J)
m = mass (g)
c = heat capacity (usually and in your case, 4.18)
^T = change in temperature
then use: ^H = Energy lost (J) / Moles
using the answer you calculated for energy (J), to get your answer in J/mol
