Exam (elaborations)
PORTAGE LEARNING CHEM 104 MODULE 1 –MODULE 6 EXAM Compiled
- Course
- CHEM 104
- Institution
- Portage Learning
PORTAGE LEARNING CHEM 104 MODULE 1 –MODULE 6 EXAM Compiled
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PORTAGE LEARNING CHEM 104x x x
xMODULE 1 –MODULE 6 EXAM
x x x x
David.jamin19@gmail.com
,PORTAGE LEARNING CHEM 104
x x x
MODULE 1 –MODULE 6 EXAM
x x x x
, Portage Learning CHEM 103 Module 1 Exam:
x x x x x x
Question x1
In xthe xreaction xof xgaseous xN2O5 xto xyield xNO2 xgas xand xO2 xgas xas xshown xbelow, xthe xfollowing
xdata xtable xis xobtained:
2 xN2O5 → x 4 xNO2 x(g) x x
+ x x O2 x(g)
x(g)
Data xTable x#2
Time x(sec) [N2O5] [O2]
0 0.300 xM 0
300 0.272 xM 0.014 xM
600 0.224 xM 0.038 xM
900 0.204 xM 0.048 xM
1200 0.186 xM 0.057 xM
1800 0.156 xM 0.072 xM
2400 0.134 xM 0.083 xM
3000 0.120 xM 0.090 xM
1. Using xthe x[O2] xdata xfrom xthe xtable, xshow xthe xcalculation xof xthe xinstantaneous xrate
xearly xin xthe xreaction x(0 xsecs xto x300 xsec).
2. Using xthe x[O2] xdata xfrom xthe xtable, xshow xthe xcalculation xof xthe xinstantaneous xrate xlate xin
xthe xreaction x(2400 xsecs xto x3000 xsecs).
3. Explain xthe xrelative xvalues xof xthe xearly xinstantaneous xrate xand xthe xlate xinstantaneous xrate.
Your xAnswer:
1. rate x= x(0.014 x- x0) x/ x(300 x- x0) x= x4.67 xx x10-5 xmol/Ls
2. rate x= x(0.090 x- x0.083) x/ x(3000 x- x2400) x= x1.167 xx x10-5 xmol/Ls
3. The xlate xinstantaneous xrate xis xsmaller xthan xthe xearly xinstantaneous xrate.
, Question x2
The xfollowing xrate xdata xwas xobtained xfor xthe xhypothetical xreaction: x A x x + x x B x x → x x X x x + x Y
Experiment x# [A] [B] rate
1 0.50 0.50 2.0
2 1.00 0.50 8.0
3 1.00 1.00 64.0
1. Determine xthe xreaction xorder xwith xrespect xto x[A].
2. Determine xthe xreaction xorder xwith xrespect xto x[B].
3. Write xthe xrate xlaw xin xthe xform xrate x= xk x[A]n x[B]m x(filling xin xthe xcorrect xexponents).
4. Show xthe xcalculation xof xthe xrate xconstant, xk.
Your xAnswer:
rate x= xk x[A]x x [B]y
rate x1 x/ xrate x2 x= xk x[0.50]x x[0.50]y x/ xk x[1.00]x x [0.50]y
2.0 x/ x8.0 x= x[0.50]x x / x[1.00]x
0.25 x= x0.5x
x x= x2
rate x2 x/ xrate x3 x= xk x[1.00]x x[0.50]y x/ xk x[1.00]x x [1.00]y
8.0 x/ x64.0 x= x[0.50]y x/ x[1.00]y
0.125 x= x0.5y
y x= x3
rate x= xk x[A]2 x[B]3
2.0 x= xk x[0.50]2 x[0.50]3
k x= x64
Question x3
ln x[A] x- xln x[A]0 x = x- xk xt 0.693 x= xk xt1/2
An xancient xsample xof xpaper xwas xfound xto xcontain x19.8 x% x14C xcontent xas xcompared xto xa
xpresent-day xsample. xThe xt1/2 xfor x
14C xis x5720 xyrs. xShow xthe xcalculation xof xthe xdecay xconstant
x(k) xand xthe xage xof xthe xpaper.
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