Question

The combustion of 1.881 g of glucose, C 6 H 12 O 6 ( s ) , in a bomb calorimeter with a heat capacity of 4.90 kJ/°C results in an increase in the temperature of the calorimeter and its contents from 22.06 °C to 28.04 °C. What is the internal energy change, Δ U , for the combustion of 1.881 g of glucose?

Answer 1

Answer:

ΔU = - 2804.02 kJ/mol

Explanation:

The equation for the reaction is:

C₆H₁₂O₆(s)    +    6O₂ (g)      ⇒          6CO₂(g)      +     6H₂O(l)    

Given that:

Calorimeter heat capacity $(q_{cal}) =4.90kJ/^0C$

change in temperature =  $(\delta T) = 28.04^0C-22.06^0C\\(\delta T) = 5.98^0C$

mass of glucose = 1.881 g

molar mass of glucose = 180 g/mol

number of moles of glucose = $\frac{mass of glucose}{molar mass of glucose}$

number of moles of glucose = $\frac{1.881g}{180g/mol}$

number of moles of glucose = 0.01045 mole

The heat absorbed (q) by the calorimeter is calculated as follows:

Heat absorbed = Heat capacity × change in temperature

$\delta H = q* \delta T$

= 4.90 × 5.98°C

= 29.302 kJ

Enthalpy of combustion can be calculated as:

$\delta H_c=-\frac{q}{n}$

where; n = number of moles of glucose = 0.01045 mole

$\delta H_c=-\frac{29.032kJ}{0.01045mol}$

$\delta H_c=- 2804.019 kJ/mol$

ΔH = - 2804.02 kJ/mol

The change in internal energy ΔU can be calculated as:

ΔH = ΔU + ΔnRT

where Δn = difference in number of moles of species in of product minus that of reactant; in that case, we have:

(6 - 6) = 0

ΔH = ΔU + ΔnRT

- 2804.02 kJ/mol = ΔU  +  (0 × R × T)

- 2804.02 kJ/mol = ΔU  + 0

- ΔU = 2804.02 kJ/mol

ΔU = - 2804.02 kJ/mol

Answer 2

Answer:

2804.02 kJ/mol.

Explanation:

Equation of reaction:

C6H12O6 + 6O2 --> 6CO2(g) + 6H2O(g)

Molar mass of C6H12O6 = (6*12) + (12*1) + (6*16)

= 180 g/mol

Number of moles = mass/molar mass

= 1.881/180

= 0.01045 mol.

ΔU = q + W

W = 0,

ΔU = q

ΔT = 28.04 - 22.06

= 5.98 °C

qrxn = qcal;

qcal = Ccal * ΔT

= 4.9 * 5.98

= 29.302 kJ

ΔU = qcal/n

= 29.302/0.01045

= 2804.02 kJ/mol.