A0.500 g sample of tnt is burned in a bomb calorimeter containing 610 g of water at an initial temperature of 20.0 celcius. the heat capacity of the calorimeter is 420 j/c and the heat of combustion of tnt is 3374 kj/mol.

(a) write and balance the chemical equation

(b) using these data, calculate the final temperature of the water and calorimeter once the reaction is complete

a) 2C₇H₅N₃O₆ + 7O₂ → 3N₂ + 5H₂O + 7CO + 7CO₂

b) 22.50°C

Explanation:

a) TNT has as molecular formula C₇H₅N₃O₆, and the explosion or combustion reaction produces nitrogen gas, water, carbon monoxide, and carbon dioxide as:

C₇H₅N₃O₆ + O₂ → N₂ + H₂O + CO + CO₂

To balance it, the number of elements must be the same on both sides of the equation. So, first, let's multiply each carbon product by 7, and C₇H₅N₃O₆ by 2:

2C₇H₅N₃O₆ + O₂ → N₂ + H₂O + 7CO + 7CO₂

Thus, N₂ must be multiplied by 3, and H₂O by 5:

2C₇H₅N₃O₆ + O₂ → 3N₂ + 5H₂O + 7CO + 7CO₂

And now O₂ by 7:

2C₇H₅N₃O₆ + 7O₂ → 3N₂ + 5H₂O + 7CO + 7CO₂

b) The calorimeter is an equipment used to calculate the energy released by combustion, and it can be considered isolated. By energy conservation:

Qcombustion + Qcalorimeter + Qwater = 0

The energy released of the combustion must be negative, the signal indicates that the heat is lost. The molar mass of TNT is 227.13 g/mol, thus the number of moles is:

n = mass/molar mass

n = 0.500/227.13 = 2.20x10⁻³ mol

So,

Qcombustion = -3374 kJ/mol * 2.20x10⁻³ mol

Qcombustion = -7.4275 kJ = -7427.5 J

Qcalorimter = C*ΔT, where C is the heat capacity, and ΔT is the temperature variation (final - initial).

Qwater = m*c*ΔT, where m is the mass, and c is the specific heat (4.184 J/g°C).

Thus,

-7427.5 + 420*ΔT + 610*4.184*ΔT = 0

(420 + 2552.24)ΔT = 7427.5

ΔT = 2.5

Tfinal - 20.0 = 2.5

Tfinal = 22.50°C

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

the answer is c. nacl