Question

Phthalates used as plasticizers in rubber and plastic products are believed to act as hormone mimics in humans. The value of ΔHcomb for dimethylphthalate (C10H10O4) is –4685 kJ/mol. Assume 0.905 g of dimethylphthalate is combusted in a calorimeter whose heat capacity (Ccalorimeter) is 6.15 kJ/°C at 21.5 °C. What is the final temperature of the calorimeter?

Answer 1

Answer:

$T_f$ = 25.05°C

Explanation:

Given:

the value of ΔHcomb (heat of combustion) for dimethylphthalate (C10H10O4) is = 4685 kJ/mol.

mass = 0.905g of dimethylphthalate

molar mass = 194.18g dimethylphthalate

number of moles of dimethylphthalate = ???

$T_i$ = 21.5°C

$C_{calorimeter}$ = 6.15 kJ/°C

$T_f$ = ???

since we have our molar mass and mass of dimethylphthalate ;we can determine the number of moles as;

0.905g of dimethylphthalate ×  $\frac{1 mole (dimethylphthalate)}{194.184g(dimethylphthalate)}$

number of moles of dimethylphthalate = 0.000466 moles

Heat released = moles of dimethylphthalate × heat of combustion

=  0.000466 moles × 4685 kJ

= 21.84 kJ

∴ Heat absorbed by the calorimeter =  $C_{calorimeter}$ $(T_f-T_i} )$

21.84 kJ =6.15 kJ/°C $* (T_f-21,5^0C)$

21.84 KJ = $(6.15 kJ/^0C * T_f) - (6.15 kJ/^0C*21.5^0C)$

21.84 KJ = $(6.15 kJ/^0C * T_f)$ - 132.225 kJ

21.84 KJ + 132.225 kJ = $(6.15 kJ/^0C * T_f)$

154.065 kJ = $(6.15 kJ/^0C * T_f)$

$T_f$ = $\frac{154.065kJ}{6.15kJ/^0C}$

$T_f$ =25.05°C

Answer 2

Answer:

The final temperature of calorimeter in combustion of dimethylphthalate is $\rm 25.05 ^0 C$.

Explanation:

The heat of combustion of dimethylphthalate is given 4685 kJ/mol.

The mass of dimethylphthalate to be combusted is 0.905 g.

We know that,

The molecular mass of dimethylphthalate is 194.18 g.

Therefore,

the moles of dimethylphthalate to be combusted is:

number of moles = $\frac{\rm weight}{\rm molecular weight}$

number of moles of dimethylphthalate =  $\frac{\rm0.905}{\rm194.18}$

number of moles of dimethylphthalate = 0.000466 moles

For a reaction, heat releases can be calculated as:

Heat release = moles of compound * heat of combustion

Heat release for dimethylphthalate =  0.000466 moles * 4685 kJ/mol

Heat release for dimethylphthalate = 21.84 kJ

For, the calculation of heat absorbed by calorimeter,

The heat absorbed = Heat capacity of calorimeter (Final temperature - Initial temperature)

Heat absorbed = $\RM 6.15 kJ/^0C\; \left ( Final temperature - 21.5 ^0 C \right )$

Heat absorbed = $\RM \left ( 6.15 kJ/^0C * Final temperature \right ) - \RM \left ( 6.15 kJ/^0C\; *\;21.5 ^0 C \right )$

21.84 kJ (from the calculation) = $\RM \left ( 6.15 kJ/^0C * Final temperature \right ) - \RM \left ( 6.15 kJ/^0C\; *\;21.5 ^0 C \right )$

21.84 kJ = $\RM \left ( 6.15 kJ/^0C * Final temperature \right )$ - 132.225 kJ

21.84 kJ + 132.225 kJ = $\RM \left ( 6.15 kJ/^0C * Final temperature \right )$

154.065 kJ = $\RM \left ( 6.15 kJ/^0C * Final temperature \right )$

$\frac{154.065\; kJ}{6.15\; kJ/^0 C}$ = Final Temperature of Calorimeter

$\rm 25.05 ^0 C$ = Final temperature of calorimeter.

For more information, refer the link:

https://brainly.com/question/15117038?referrer=searchResults