CHM 139 2011 April answer key

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!
Multiple!Choice!
1. The!amount!of!A!is!reduced!to!15%!of!the!original!but!B!is!only!
reduced!to!17%.!!This!would!mean!that!A!is!the!limiting!reagent,!
and!therefore!the!amount!of!product!would!also!be!reduced!to!
15%…500!g!x!0.15!=!75!g!
!@).E,-(3(
!
2. The!height!of!the!liquid!is!higher!on!the!atmospheric!side,!so!
Pgas>Patm!and!Pgas!=!Patm!+!Ph!
!@).E,-(3(
!
3. 3s!orbital!has!2!spherical!nodes!
!@).E,-(F(
!
4. 3RT/MMSF6!=!3RT/MMAr(
!(3)(8.314)(T)/146.053!g/mol!=!(3)(8.314)(293!K)/39.948!g/mol!
!T!=!1071.4!K!=!798oC(
!@).E,-(F(
!
5. Normal!boiling!point!occurs!when!vapour!pressure!equals!
atmospheric!pressure!(1!atm!=!760!mm!Hg)!
!@).E,-(F(
(
6. As!IMF!increases,!fewer!molecules!escape!from!a!liquid!into!the!
vapour!phase!(VP)!
!@).E,-(<(
!
7. @).E,-!G(
!
8. Normal!boiling!point!is!VP!=!Patm!so!Pmixture!=!760!mm!Hg!at!80oC.(
! PA!+!PB!=!760!mm!Hg!
! PA!=!PA
o!XA!=!(700!mm!Hg)!XA!
! PB!=!PB
o!XB!=!PB
o!(1!!XA)!=!(940!mm!Hg)(1!!XA)!
(700!XA!+!940(1!!XA)!=!760!and!XA!=!0.75!
!@).E,-(G(
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!"#$%&"'!(
!
9. ln!([A]/[A]o)!=!kt!=!!(0.00073/s)(300!s)!=!0.219!
![A]/[A]o!=!e0.219!=!0.803!=!80!%!
!@).E,-(@(
!
10. This!is!the!definition!of!a!reaction!that!is!zeroorder!with!respect!to!
reactants.!
!@).E,-(<(
!
11. The!rate!law!is!written!with!respect!to!the!reactants!of!the!rate
determining!step!(the!slow!step)!
(@).E,-(F!
!
12. The!top!reaction!is!reversed,!the!bottom!reaction!is!doubled(
! Kc!=!(1/0.0667)(4)2!=!240!
! Kp!=!Kc!(RT)Δn!=!(240)[(0.0821)(298)]‐5!=!2.74!x!10‐5(
!@).E,-(F(
!
13. !
!
!
!
!(Y!!2x)!+!2x!+!0.3500!=!1.0866!(from!law!of!partial!pressures)!
!Y!=!0.7366!
! Kp!=!(NO)2(O2)/(NO2)2!=!(2!x!0.3500)2(0.3500)/[0.7366(2)(0.3500)]!
!=!128(
!@).E,-(G(
(
14. At!the!halfequivalence!point,!pH!=!pKa!
!pKb!=!log(3.7x10‐4)!=!3.43,!and!pKa!=!14!!pKb!=!10.57!=!pH!@!½!eq!
!@).E,-(<(
!
15. Ksp!=!4x3!=!1.5!x!1010(
!x!=!3.35!x!10‐4!mol/L!3.35!x!10‐4!mol/L!x!78!g/mol!=!0.026!g/L(
!@).E,-(G(
!
!
2NO2 -->
2NO +
O2
I Y 0 0
C - 2x + 2x + x
E Y - 2x 2x 0.3500
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