Please give answers to these other questions too. (this is not a duplicate post.
ID: 880708 • Letter: P
Question
Please give answers to these other questions too. (this is not a duplicate post. I split the assingment into 2)
All work must be shown :)
Graph 1 Graph 2 15 15 Auto Fit for Inverse Functions (Student) | Moles of gas n = A/T+B A: 3389 +- 142.0 B: -1.305 +/-0.4979 RMSE: 0.1924 2 13 13 3 11 9 Linear Fit for: Inverse Functions (Student) I Moles of gas n = mx+ b m (Slope): 3522 mmol/1/K b (Y-Intercept): -1.870 mmol Correlation: 0.9988 RMSE: 0.06838 o ] 8 7 7 0.0040 0.0035 1T (1/K) 250 300 350 0,0030 Temperature (K)Explanation / Answer
Graph 1 and 2
Slope=3522 mmol/K
Intercept= -1.370 mmol
n( no of milll moles)= 3522(mmol/T)*T(K)-1.870mmol
Graph 3 and 4
slope is -0.01257 atm/min ( on log log sclae)
k= 0.01284 min-1
b==exp(-0;4054) =0.;66671 atm
P0= 0.6792 atms
Equation is lnP= -0.01257(atm/min)*t+0.4054 atma
Pressure at 10 minutes
P= 0.6792*exp(-0.01284*10) =0.6792*0.8795= 0.59 atma
after 20 hrs , 1200 minutes
P=0.6792*exp(-0.01284*1200) =1.382X10-7 atma= 1.382X10-7*760 torr= 0.0001053torr
time taken for 250 torr = 250/760 atma = 0.3289 atma
ln(250/760)= =-.01257*t+ 0.4054
-1.11186= -0.01257t+0.4054
-1,51726 =-0.01257t
t= 120 minutes =2 hrs
Initial pressure can be obtained from P =Po exp(-kt)
put t= o
P= 0.6792 *1 =0.6792 atma =0.6792 atma =P0
P at half life= 0.6792/2= 0.3396
from P= Poexp(-kt)
0.3396= 0.6792 exp(-kt)
1/2 =exp(-kt)
ln(1/2)= -kt1/2
0.6932 = k t1/2
t1/2= 0.693/ k= 0.693/ 0.01284 =53.98 minutes
when the inititla pressure is 0.2 atma. the half life remains the same since the reaction is first oder equation and depends only on K.
ln250-