I have ONE question for homework that is separated into different parts. 1) a) A

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) H = -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) H= +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/(gC) for the solution, an initial temperature of 26.0 C, and no heat transfer between the cold pack and the environment.

d)Calculate Hf (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) H = -6534kJ
Hf (CO2) = -393.5kJ/mol
Hf (H2O) = - 285.8kJ/mol

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

1) H° to be negative and S° to be negative or 2) H° to be positvie and S° to be positive.

Explanation / Answer

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) H = -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.

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