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The combustion of 1.774 g of glucose, C6H12O6(s), in a bomb calorimeter with a heat capacity of 4.20 kJ/°C results in an increase in the temperature of the calorimeter and its contents from 23.00 °C to 29.58 °C. What is the internal energy change, Δ????, for the combustion of 1.774 g of glucose?

Respuesta :

Answer:

Answer : 2812.6 KJ/mol

Explanation:

We know that molar mass of glucose = 180 g/mol

[tex]Number\ of\ mole =\frac{ (given\ mass)}{(molar\ mass)}[/tex]

[tex]= \frac{(1.774}{180)} mole[/tex]

= 0.00985 mole

[tex](T_2 - T_1)[/tex] has same value in degree C or in Kelvin

[tex]q = s\times (T2-T1)[/tex]

[tex]= 4.20\times (29.58 - 23.00)[/tex]

       = 27.636 KJ

for 0.00985 mole, heat = 27.636 KJ

for 1 mole heat [tex]= \frac{27.636}{0.00985} KJ/mol[/tex]

= 2805.6 KJ/mol

so,

[tex]\Delta E[/tex] for the combustion of glucose = 2805.6 KJ/mol

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