Question

1. Given the following:

C_(s) + 2H_2(g) → CH_4(g)                         ΔH= −74.6 kJ

C_(s) + 2Cl_2(g) → CCl_4(g)                       ΔH= −95.7 kJ

H_2(g) + Cl_2(g) → 2HCl_(g)                      ΔH= −184.6 kJ

Calculate the enthalpy of reaction for:

CH_4(g) + 4Cl_2(g) → CCl_4(g) + 4HCl_(g)

2. There was a lot of talk among football fans about "deflate-gate" last week (hopefully it's not still going on this week) where the New England Patriots grounds crew seem to have decreased the pressure in the team balls by around 2 PSI from the official value of 12 PSI. (These are Gauge pressures, meaning pressure above 1 atm.) A Physics prof from Boston U. suggested that it may not have been a nefarious plot, but simply a consequence of the fact that it's cold outdoors in Boston and they may have filled the balls indoors. See if this makes sense by calculating the change in pressure (assuming the football had constant volume) between 70 degrees F (nominal inside temperature) and 20 degrees F (a relatively balmy winter-time outside temperature in Boston) from the assumed starting point of 12 PSI.

The Fahrenheit to Celsius conversion is ºC = (ºF – 32)/1.8, there are 14.7 PSI in 1.0 atm, and the volume of a football is 4237 cm³ – though you may not even need that last one.

3.  Imagine you were in the mountains and were desperate for a cup of coffee (or tea, if that's your thing) and the only source of heat you have is a butane cigarette lighter. Could you heat a cup of water at 25 ºC to near boiling with the full capacity of the lighter?

You'll have to make some assumptions about the amount of butane in the lighter and water in your cup, which you should identify; but FYI the density of liquid butane is about 0.6 g/mL, the MW is 58.12 g/mol, and the heat of combustion is -2623 kJ/mol (and of course, the density of water is 1.0 g/mL and the heat capacity is 4.18 J/ºC-g).

4.

Answer 1

The Fahrenheit to Celsius conversion is ºC = (ºF – 32)/1.8, there are 14.7 PSI in 1.0 atm, and the volume of a football is 4237 cm³ – though you may not even need that last one.

At 70 deg.F

P1/T1= P2/T2

P2= P1*T2/T1

T1= 70 deg.F = (70-32)/1.8= 21.11deg.c= 21.11+273.15= 294.2611 K

P1= (1atm+12/14.7 atm)= 1.816 atm

P2= P1*T2/T1

deg.F	deg.C	deg.K (T2)	T2/T1	P2 (atm)	P1(PSI)
70	21.11111	294.2611	1	1.816	11.9952
60	15.55556	288.7056	0.98112	1.781715	11.4912
50	10	283.15	0.980757	1.781055	11.4815
40	4.444444	277.5944	0.980379	1.780369	11.47143
30	-1.11111	272.0389	0.979987	1.779656	11.46094
20	-6.66667	266.4833	0.979578	1.778914	11.45003

1 mol of butane produces 2623 Kj =

Let 100ml be the amount of water

Mass of water= 100*1 =100 gm

Volume of butane= 1ml (assumed)

Mass of butane= 1*0.6 =0.6gms

Moles of butane= 0.6/58.12=0.010323

Heat produced= 0.010323* 2623*1000 Joules=27078.46

This is used to boil water

27078.46= mass* specific heat* temperature difference= 100*4.18*(t-25)

T= 27078.46/ 418=67.41 deg.c

So it is not possible to boil water with 1ml of liquid butane