Question

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?

Answer 1

Answer:

Answer : 2812.6 KJ/mol

Explanation:

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

$Number\ of\ mole =\frac{ (given\ mass)}{(molar\ mass)}$

$= \frac{(1.774}{180)} mole$

= 0.00985 mole

$(T_2 - T_1)$ has same value in degree C or in Kelvin

$q = s\times (T2-T1)$

$= 4.20\times (29.58 - 23.00)$

       = 27.636 KJ

for 0.00985 mole, heat = 27.636 KJ

for 1 mole heat $= \frac{27.636}{0.00985} KJ/mol$

= 2805.6 KJ/mol

so,

$\Delta E$ for the combustion of glucose = 2805.6 KJ/mol