temp. ti = 25.0 o C Final temp, tf = 47.00 o C , heat capacity of calorimeter =

Question

temp. ti = 25.0oC

Final temp, tf = 47.00oC , heat capacity of calorimeter = 9.84 kJ/K

First we need to calculate the heat absorbed by calorimeter and then from that we need to calculate the heat given by Ti. The calculating the moles of 0.515 g of Ti and then need to convert the energy from J to J/mol and kJ/mol.

We know,

Heat gain by calorimeter = Heat loss by Ti

Heat gain by calorimeter

q = heat capacity C * t

Heat capacity in kJ/K so need to convert the J/K

1 kJ = 1000 J

So, 9.84 kJ = ?

= 9840 J/K

q = heat capacity C * t

    = 9840 J/K * (47.00-25.0)oC

    = 216480 J

We Heat gain by calorimeter = Heat loss by Ti

So, Heat loss by Ti = - 216480 J

                                = -216.48 kJ

Now moles of Ti = 0.515 g / 47.87 g.mol-1

                             = 0.010758 moles

So, 0.010758 moles of Ti = -216.48 kJ

So, 1 moles of Ti = - 2.00 *10^4 kJ/ mol

So, the heat of the reaction for the combustion of the Ti in this calorimeter is -2.00*104 kJ/mol

Explanation / Answer

The combustion of titanium with oxygen produces titanium dioxide:

Ti(s) + O2(g) TiO2(s)

When 0.515 g of titanium is combusted in a bomb calorimeter, the temperature of the calorimeter increases from 25.00°C to 47.00°C. In a separate experiment, the heat capacity of the calorimeter is measured to be 9.84 kJ/K. The heat of reaction for the combustion of a mole of Ti in this calorimeter is ________ kJ/mol.

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