Two experiments were conducted in a bomb calorimeter. The first one to determine the heat capacity of the calorimeter, the second the heat of combustion of the carcinogenic substance benzene (C6H6). a. In the first experiment, the temperature rises from 22.37 o C to 24.68 o C when the calorimeter absorbs 5682 J of heat. Determine the heat capacity of the calorimeter. Page 3 of 4 b. In the second experiment, the combustion of 0.258 g of benzene increases the temperature from 22.37 o C to 26.77 o C. Determine the heat of combustion for 1 mol of benzene.

The right solution is:

(a) 2459.74 J/degree C

(b) 3271.769 KJ/moles

Explanation:

According to the question,

(a)

The heat capacity of the calorimeter will be:

=

=

=

(b)

The change in temperature will be:

=

=

The amount of heat released will be:

=

=

or,

=

Moles of benzene combusted will be:

=

=

hence,

The heat combustion for 1 mol of benzene will be:

=

=

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol