A 0.500 g sample of C7H5N2O6 is burned in a calorimeter containing 600. g of water at 20.0∘C. If the heat capacity of the bomb c

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A 0.500 g sample of C7H5N2O6 is burned in a calorimeter containing 600. g of water at 20.0∘C. If the heat capacity of the bomb c

alorimeter is 420.J∘C and the heat of combustion at constant volume of the sample is −3374kJmol, calculate the final temperature of the reaction in Celsius. The specific heat capacity of water is 4.184 Jg ∘C.

Chemistry

1 answer:

lorasvet [2.7K] · Answer 1 · 2026-03-25T15:24:15+03:00

Answer:

22.7

Explanation:

Initially, calculate the energy released from the sample's mass. The combustion heat represents the energy per mole of the fuel:

ΔHC=qrxnn

We can rearrange this formula to isolate qrxn, remembering to convert the sample's mass into moles:

qrxn=ΔHrxn×n=−3374 kJ/mol×(0.500 g×1 mol213.125 g)=−7.916 kJ=−7916 J

The heat released during the reaction must match the total heat absorbed by both the water and the bomb calorimeter:

qrxn=−(qwater+qbomb)

The heat absorption by the water can be calculated using its specific heat:

qwater=mcΔT

The calorimeter's heat absorption can be derived from its heat capacity:

qbomb=CΔT

Combine both equations into the first equation, substituting the known details, with ΔT=Tfinal−20.0∘C:

−7916 J=−[(4.184 Jg ∘C)(600. g)(Tfinal–20.0∘C)+(420. J∘C)(Tfinal–20.0∘C)]

Distribute each multiplication term and simplify:

−7916 J=−[(2510.4 J∘C×Tfinal)−(2510.4 J∘C×20.0∘C)+(420. J∘C×Tfinal)−(420. J∘C×20.0∘C)]=−[(2510.4 J∘C×Tfinal)−50208 J+(420. J∘C×Tfinal)−8400 J]

Combine the similar terms and simplify:

−7916 J=−2930.4 J∘C×Tfinal+58608 J

Finally, isolate Tfinal:

−66524 J=−2930.4 J∘C×Tfinal

Tfinal=22.701∘C

Round to three significant figures gives the final result as 22.7∘C.