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At constant volume, the heat of combustion of a particular compound, compound A, is –3568.0 kJ/mol. When 1.411 g of compound A (molar mass = 115.27 g/mol) was burned in a bomb calorimeter, the temperature of the calorimeter (including its contents) rose by 3.379 °
c. Using this data, what is the heat capacity (calorimeter constant) of the calorimeter?

Respuesta :

Answer : The heat capacity (calorimeter constant) of the calorimeter is, [tex]12.92kJ/^oC[/tex]

Explanation:

First we have to calculate the moles of compound A.

[tex]\text{Moles of compound A}=\frac{\text{Mass of compound A}}{\text{Molar mass of compound A}}[/tex]

[tex]\text{Moles of compound A}=\frac{1.411g}{115.27g/mol}=0.01224mol[/tex]

Now we have to calculate the heat of combustion of compound A for 0.01224 mol.

As, 1 mole of compound A has heat of combustion = 3568.0 kJ

So, 0.01224 mole of compound A has heat of combustion = 0.01224 × 3568.0 kJ

                                                                                                 = 43.67 kJ

Now we have to calculate the heat capacity (calorimeter constant) of the calorimeter.

[tex]q=c\times \Delta T[/tex]

where,

q = heat of combustion = 43.67 kJ

c = heat capacity = ?

[tex]\Delta T[/tex] = change in temperature = [tex]3.379^oC[/tex]

Now put all the given values in the above expression, we get:

[tex]43.67kJ=c\times (3.379^oC)[/tex]

[tex]c=12.92kJ/^oC[/tex]

Therefore, the heat capacity (calorimeter constant) of the calorimeter is, [tex]12.92kJ/^oC[/tex]

A device that is used for measuring the heat and heat capacity formed during the reaction involving chemical, electrical and mechanical modes is called a calorimeter.

The heat capacity of the calorimeter is [tex]12.92 \rm \;kJ/\°C[/tex]

How to calculate the heat capacity?

Given,

  • The heat of combustion of compound A (q)= [tex]- 3568.0 \;\rm kJ/mol[/tex]
  • Weight of the compound A (m)= 1.411 g
  • The molar mass of the compound A (M) = [tex]115.27 \;\rm g/mol[/tex]
  • Change in temperature =[tex]3.379 \;\rm \°C[/tex]

Step 1: Calculate the moles of the compound A:

[tex]\begin{aligned}\rm Moles (n) &= \dfrac{\rm m}{\rm M}\\\\\rm n &= \dfrac{1.411 \;\rm g}{115.27\;\rm g/mol}\\\\\rm n &= 0.01224 \;\rm mol\end{aligned}[/tex]

Step 2: Calculate the heat of combustion for compound A:

1 mole of compound = 3568.0 kJ of the heat of combustion

So, 0.01224 mole of compound A = ?

[tex]\begin{aligned}q &= 0.01224 \times 3568.0 \;\rm kJ\\\\q&= 43.67 \;\rm kJ\end{aligned}[/tex]

Step 3:Calculate the heat capacity of the calorimeter:

[tex]\begin{aligned}\rm q &= \rm c \times \Delta T\\\\43.67 &= \rm c \times 3.379\\\\\rm c &= 12.92 \;\rm kJ/ \°C\end{aligned}[/tex]

Therefore, the heat capacity of the calorimeter is [tex]12.92 \rm \;kJ/\°C[/tex].

Learn more about heat capacity and calorimeter here:

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