Part A: My driveway is 50 feet long and 10 feet wide. My driveway is coated in i
ID: 1020409 • Letter: P
Question
Part A: My driveway is 50 feet long and 10 feet wide. My driveway is coated in ice to an average depth of 1.2 inches. The current temperature is -5 ºC. How much ethanol (g) would I need to throw on the driveway to have an ice free driveway?
2: Looking at Experiment #2 and #3 in the data table, answer the following questions:
Expt # Initial Nitrogen (atm) Initial Chlorine (atm) Initial Rate (atm/min) Temp (K)
1 0.250 0.250 0.0115 300 K
2 0.500 0.500 0.045 300 K
3 0.500 0.250 0.011 300 K
4 0.250 0.500 0.072 500K
A. What is the ratio of the initial concentration of nitrogen in Experiment #2 relative to Experiment #3?
B. What is the ratio of the intial concentration of chlorine in Experiment #2 relative to Experiment #3?
C. What is the ratio of the rate of the reaction in Experiment #2 relative to Experiment #3?
D. Look back at your answers to Piece #4. Based on your answers to Part C and Part B, what is the order of the reaction with respect to chlorine?
E.What is the order of the reaction with respect to nitrogen?
F. Write the rate law for the reaction.
G. What is the rate constant (k) for the reaction at 300 K, with appropriate units?
H.What is the rate constant (k) for the reaction at 500 K, with appropriate units?
I. 1 atm of nitrogen and 1 atm of chlorine are mixed in a 2 L flask at 500 K. What is the initial rate of the reaction?
Explanation / Answer
1.
This much ice at -5 deg.c t need to be melted by ethanol. Ice needs to be heated from -5 deg,c to 0 deg.c and at 0 deg.c it needs to be melted at constant temperautre
Sensbie heat of ice= mass* specific heat of ice* (0+5)= 1326*2.05*5=13592 joules
Latent heat of fusion= 334 joules/gm
Heat required to melt the ice = 334* 1326 =442884 joules
Total heat to be suppled = 13592+442884 =456476 joules
This has to come from evaporation of ethanol whose latent heat is 40.65*1000 joules/mole
Moles of ethanol required = 456476/(40.65*1000)=11.23 moles
Mass of ethanol = moles* molecular weight= 11.23*46=516. 6 gms
a)
Initial concentration of nitrogen in experiment-3= 0.5
Ratio =0.5/0.5= 1
b). 0.5/0.25= 2
c)0.045/0.011= 4.1
d)
From experiment-2 and 3, while keeping the nitrogen concentration constant, when the concentration of Cl2 is halved, the rate reduced by 4 times. The rate with respect to Chlorine is (0.045/0.011)1/2 =2
e)
The rate has not changed from experiment 1 and 3 even though N2 concentration has been changed from 0.25 atm to 0.5 without change in Chlorine concentration . So the reaction rate is independent of nitrogen concentration and hence zero order with respect to N2.
f) The rate expression -r = K[Cl2]2
g) From experiment, 1
K[0.25]2 =0.0115, K = 0.184/atm.min
h) from experiment-4
At 500k, K[0.5]2 =0.072, K =0.288/atm.nin
For 1atm N2 and1 atm chlorine, partial pressure of Cl2= ½= 0.5
Rate= 0.288*[0.5]2 =0.072 atm/min