Question

Please show work for the three problems... I have the answers but don't know how to do the problems. Please show each one step by step

1.)

Determine the heat of reaction for the combustion of ammonia 4NH3(g)+702(g)-»4NO2(g) + 6H20() using molar enthalpies of formation AH° (kJ/mol) 45.9 molecule NH3(g) NO2(g) H2O() +33.1 -285.8 a. +30.24 kJ b. -206.9 kJ c. -298.6 kJ d. -1398.8 kJ e. -1663.6 k

Answer: D

2.)

Answer: A

3.)

Answer: D

Answer 1

1. Heat of a reaction (Δ H) = ∑ n Δ H (products) - ∑ n Δ H (reactants)

4 NH₃ (g) + 7 O₂ (g) 4 NO₂ (g) + 6 H₂O (l)

n = 4 for NH₃, 7 for O₂, 4 for NO₂, 6 for H₂O

∑ n Δ H (products) = 4 mol of NO₂ (33.1 kJ/mol)+ 6 mol of H₂O (-285.8 kJ/mol)

∑ n Δ H (products) = +132.4 kJ/mol-1714.8 kJ/mol = -1582.4 kJ/mol

∑ n Δ H (reactants) = 4 mol of NH₃ (-45.9 kJ/mol) + 7 mol of O₂ (0 kJ/mol)

∑ n Δ H (reactants) = -183.6 kJ/mol

Therefore, Heat of a reaction (Δ H) = -1582.4 kJ/mol – [-183.6 kJ/mol]

Therefore, Heat of a reaction (Δ H) = -1582.4 kJ/mol + 183.6 kJ/mol = -1398.8 kJ/mol

2. Heat of the reaction for the decomposition of benzene to acetylene

C₆H₆ (l)        3 C₂H₂ (g)⸱⸱⸱⸱⸱⸱⸱ ⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱⸱(1)

Here n= 1 for C₆H₆ and 3 for C₂H₂

Heat of a reaction (Δ H) = ∑ n Δ H (products) - ∑ n Δ H (reactants)

Given the heat of decomposition (Δ H) for 2 moles of C₆H₆ as -6271 kJ,

therefore, Δ H for 1 mole of C₆H₆ = = -3135.5 kJ

-6271 kJ

Also given the heat of decomposition of 2 moles of C₂H₂ = -2511 kJ

But in the in our equation (1) is C₂H₂ formed so heat of formation of 2 moles of C₂H₂ = +2511 kJ

Therefore heat of formation of 1 mole of C₂H₂ = = +1255.5 kJ

Heat of a reaction (Δ H) = [1 × -3135.5 kJ] – [3 × 1255.5 kJ] = -3135.5 kJ + 3766.5

Heat of a reaction (Δ H) = 631 kJ

3. Lets assume that the final temperature of the mixture is x°C

Now it is obvious that 22.7 < x < 87.5

We are supplied with the specific heat capacity of water = 4.184 J/g°C

Mass of sample cold water = 35.0 g

Temperature of cold water = 22.7 °C

Mass of sample hot water = 65.0 g

Temperature of hot water = 87.5 °C

Now we know that here,

energy lost by hot water = energy gained by cold water

(Mass of hot water) × Spec.heat capacity × (Temp. of hot water - x) =

(Mass of cold water) ×   Spec.heat capacity × (x - Temp. of cold water)

Substituting the given values,

65.0 g × 4.184 J/g°C × (87.5 - x) °C = 35.0 g × 4.184 J/g°C × (x – 22.7) °C

271.96 × (87.5 - x) = 146.44 × (x – 22.7)

23796.5 – 271.96x = 146.44x – 3324.188

146.44x + 271.96x = 23796.5 + 3324.188

418.4x = 27120.688

x = = 64.82°C

27120.688 15.