Question

Mothballs are composed primarily of the hydrocarbon naphthalene (C10H8). When 1.022 g of naphthalene burns in a bomb calorimeter, the temperature rises from 25.884 ∘C to 31.068 ∘C. (I already found ΔrU is -3323 kJ in Part A )

PART B) Find ΔrH for the combustion of naphthalene at 298 K. Express your answer using four significant figures.

2) How much heat (in kilojoules) is evolved in converting 1.00 mol of steam at 160.0 ∘C to ice at -50.0 ∘C? The heat capacity of steam is 2.01 Jg−1∘C−1 and of that ice is 2.09 Jg−1∘C−1.

Answer 1

Part B)

Here, the reaction is ,

C₁₀H₈ (s) + 12O₂ (g)   $\rightarrow$    4H₂O (l) + 10CO₂ (g)

We have,

$\triangle$H = $\triangle$U + $\triangle$nRT

Where $\triangle$n is the difference in number of moles of gaseous products and gaseous reactants = -2

R is the gas constant = 8.314 J/K/mol

Here, $\triangle$U = -3323 KJ = -3323000 J

Therefore,

$\triangle$H = -3323000 + (-2 x 8.313 x 298K) = -3327.955 KJ = -3327 KJ (Four significant figures)

2) Here, amount of heat released, Q = mcdt + mc'dt + mc"dt"+ mH_L + mH_L'             -----------------(1)

Where

dt = 160-100 = 60 ^oC

dt' = 100 -0 =100 ^oC and dt'' = 0-(-50) = 50 ^oC

H_L = Heat of vaporization of water = 2260 J/g

H_L'= Heat of fusion of ice = 334.9 J/g

m = mass of steam = 18 (g/mol) x 1 mol = 18.0 g

c = Specific heat of steam = 2.1 J/g ^oC

c' = Specific heat of water = 4.186 J/g ^oC

c" = Specific heat of ice= 2.09 J/g ^oC

Applying in equation (1),

Q = (18 x 2.1 x 60) + (18 x 4.186 x 100) + (18 x 2.09 x 50) + (18 x 2260) + (18 x 334.9) = 58392 J = 58.39 KJ