Question

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 calorimeter 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.

Answer 1

Answer:

22.7

Explanation:

First, find the energy released by the mass of the sample. The heat of combustion is the heat per mole of the fuel:

ΔHC=qrxnn

We can rearrange the equation to solve for qrxn, remembering to convert the mass of sample into moles:

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

The heat released by the reaction must be equal to the sum of the heat absorbed by the water and the calorimeter itself:

qrxn=−(qwater+qbomb)

The heat absorbed by the water can be calculated using the specific heat of water:

qwater=mcΔT

The heat absorbed by the calorimeter can be calculated from the heat capacity of the calorimeter:

qbomb=CΔT

Combine both equations into the first equation and substitute the known values, 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 the terms of each multiplication 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]

Add the like terms and simplify:

−7916 J=−2930.4 J∘C×Tfinal+58608 J

Finally, solve for Tfinal:

−66524 J=−2930.4 J∘C×Tfinal

Tfinal=22.701∘C

The answer should have three significant figures, so round to 22.7∘C.

Answer 2

          A 0.500 g sample of C7H5N2O6 is burned in a calorimeter containing 600g of water. From the given information, the final temperature of the reaction is 22.71° C

From the given parameters:

• mass of the sample C7H5N2O6 = 0.500 g

• mass of water = 600 g
• initial temperature = 20.0° C

• heat capacity of bomb calorimeter $\mathbf{c_{bc}}$ = 420 J/° C
• heat of combustion $\mathbf{\Delta H^0_c}$ = -3374 kJ/mol
• specific heat capacity of water c = 4.184 J/g° C
• TO find the final temperature = ???

From the listed parameters, the first step is to determine the number of moles of the sample by using the relation:

$\mathbf{number \ of \ moles = \dfrac{mass}{molar \ mass}}$

$\mathbf{number \ of \ moles = \dfrac{0.500 \ g}{213 \ g/mol}}$

number of moles of C7H5N2O6 = 0.00235 moles

However, the heat of combustion is the amount or quantity of heat energy released when one mole of a sample is burned.

Mathematically;

$\mathbf{\Delta H_c =\dfrac{q_{rxn} }{ n}}$

where;

$\mathbf{ q_{rxn}}$ = quantity of heat released

n = number of moles

Making  $\mathbf{ q_{rxn}}$ the subject of the formula:

$\mathbf{ q_{rxn}= \Delta H_{rxn} \times n }$

$\mathbf{ q_{rxn}=-3374 \ kJ/moles \times 0.00235 \ moles }$

$\mathbf{ q_{rxn}=-7.9289 \ kJ }$

$\mathbf{ q_{rxn}=-7928.9 \ J }$

The heat released $\mathbf{ q_{rxn}}$ = heat absorbed by bomb calorimeter + heat absorbed by water

∴

$\mathbf{ q_{rxn}= -(m\times c_{bc} \Delta T + c_{water} \times \Delta T)}$

$\mathbf{-7928.9 \ J = -( 420 J/^0C \times \Delta T + 600 \ g \times 4.18 \ J/g ^0 C\times \Delta T)}$

$\mathbf{7928.9 = (2928 \Delta T )}$

$\mathbf{ \Delta T = \dfrac{7928.9}{2928} }$

$\mathbf{ \Delta T =2 .71 ^0 \ C}$

Since ΔT = $\mathbf{T_2 - T_1}$

2.71° C = T₂ - 20° C

T₂ = 20° C + 2.71° C

T₂ = 22.71° C

Therefore, we can conclude that the final temperature of the reaction is:

22.71° C

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