Question

I have ONE question for homework that is separated into different parts.

1)

a) Assume that the kinetic energy of a 1300 kg car moving at 120 km/h could be converted entirely into heat.

What amount of water could be heated from 21 ∘C to 51 ∘C by the car's energy?

b) Regarding the water in the car how much heat (in kilojoules) is evolved or absorbed in the reaction of 2.50 g of Na with H2O?
2Na(s)+2H2O(l)→2NaOH(aq)+H2(g) ΔH∘ = -368.4kJ.

c) Instant cold packs used to treat a person injured in a car accident contain solid NH4NO3 and a pouch of water. When the pack is squeezed, the pouch breaks and the solid dissolves, lowering the temperature because of the endothermic reaction
NH4NO3(s)+H2O(l)→NH4NO3(aq) ΔH= +25.7 kJ.

What is the final temperature in a squeezed cold pack that contains 44.0 g of NH4NO3 dissolved in 125 mL of water on a person who suffered a car accident? Assume a specific heat of 4.18J/(g⋅∘C) for the solution, an initial temperature of 26.0 ∘C, and no heat transfer between the cold pack and the environment.

d)Calculate ΔH∘f (in kilojoules per mole) for benzene, C6H6,(a chemical commonly used by cars) from the following data:
2C6H6(l)+15O2(g)→12CO2(g)+6H2O(l) ΔH∘ = -6534kJ
ΔH∘f (CO2) = -393.5kJ/mol
ΔH∘f (H2O) = - 285.8kJ/mol

e) For the reaction, 2 NH₃(g) → N₂(g) + 3 H₂(g), one would expect

1) ΔH° to be negative and ΔS° to be negative or 2) ΔH° to be positvie and ΔS° to be positive.

Answer 1

Q1.

a) Assume that the kinetic energy of a 1300 kg car moving at 120 km/h could be converted entirely into heat.

What amount of water could be heated from 21 ∘C to 51 ∘C by the car's energy?

V = 120 km/h = 120 km/h * 1h/3600 s * 1000m / km = 33.3333 m/s

Ek = 1/2*m*v^2 = 1/2*(1300)(33.3333^2) = 722220.7 J

then

Q = m*C*(Tf-Ti)

722220.7 = m*4.184*(51-21)

m = 722220.7/4.184 / (30) = 5753.82grams

m = 5.753 kg of water

b) Regarding the water in the car how much heat (in kilojoules) is evolved or absorbed in the reaction of 2.50 g of Na with H2O?
2Na(s)+2H2O(l)→2NaOH(aq)+H2(g) ΔH∘ = -368.4kJ.

mol of Na = mass/MW = 2.5/22.989769 = 0.1087 mol of Na

then

2 mol --> -368.4 kJ

0.1087mol --> x

x = 0.1087/2*-368.4

x = -20.02254 kJ released

For all further questions, please consider posting them in different set of Q&A. We are only allowed to answer to 1 question per set. These questions should be posted individually, since they do not depend on each other.