物理化学二期中文试卷答案2008级

一、选择题(共10题20分) (3题35分15解：(A)因[B]0≫[A]0,rk[A][B设=1，且a (2分k=1/tln([A]0/[A])=1/20ln(1.000/0.692)=0.0184s-k0.0185,0.0177,0.0185,0.0184s-1A]0固定(同理k=0.0185,0.0167,0.0202,0.0181s-1[A]0改变，时间相对性 (4分k值相近，故Ak0.0184+0.0185+0.0177+0.0185+0.0184)/50.0183s-1A]0固定(k0.0184+0.0185+0.0167+0.0202+0.0181)/50.0184s-1A]0改变，时间相对性k=k/[B]0=0.0183/1.0=0.0183mol-1dm3s-(t1/2=ln2/kln2/0.018337.9 (3分k=k[B]0=0.01830.5=0.00915s-t1/2ln2/kln2/0.0091575.8 (3分ln(k2/k)=Ea/R(1/T-1/T2)ln(k2/0.0183)=83.6103/8.314(1/300-1/323)k2=0.199mol-1dm3s-rk2[A][B0.01990.10.10.00199mol·dm-3s- (3分题库[答]（A）对A是一级 k=0.018s-1，t1/2=39 (5分（B）t'1/2=78 (5分（C）300K时, k2=0.018mol-1·dm3·s-1 (5分)323K时, k'2=0.193mol-1·dm3·s-1,k' E 1ln2a k2 R 101

1KM

(2分

KM

r1

(1.241.160.881.28)μmoldm31.14μmoldm3

(2分r

r

kM2.35μmoldm3

(2分 kr0 0

2.35μmoldm3 0.02μmoldm3 117.5

(2分k2

117.5103dm3μmol-1s-

(2分10a答：H=E2RT65.4328.3140.3)kJmol-160.44kJmol1(2分aU=HnRT(60.448.3140.3)kJmol-162.94kJ

(2分 S=R(ln

RT

(4分 hp (-

8.314(ln(2.64109)2)JK1mol1147.6JK1G=(60.44+147.60.3)kJmol1104.7kJ

(2分(4题45分kAkA2答 (2分d[A]d

[A]2

[A

=2k

)2

([A]1

0 20=4k

(k[A]2

[A]0,2k

(4分

2

[A]

，0dln

[A]

) tt

0te14k[A]e

(4分 6．10d[CH2XCOCH3]k[CHXC(OH)CH][X

d[CH2XC(OH)CH3]0 k[CH=C(OH)CH][X

d[CH2=C(OH)CH3]0 d[CH)COH+

(3分 3 k[(CH)CO)[HA]-k [(CH)(OH+)[A-]k[(CH)(OH+)[A- 3

3 3[(CH)COH+][A]k1[(CH3)2

(2分3

1k2

r

k1

[(CH3

(1分当k2k1，rk1[(CH3)2CO][HA]反应由k1控制 (2分当

，r

k1[(CH)CO][HA]反应由k2控 (2分2k7.102k

3r=kθApB=kpBK1pA/(1+K1pA+K2pB) (6分)若A、B均为弱吸附，则K1pA+K2pB≪1，r=kK1pApB (2分)A、B均为一 (2分[答

r

p,

KA

(4分A

1KpKr

KApA1KApAKB

A B

（2分A、B

1KApAKBpB

（2分rkKApA

对A、B均为一级 （2分15（1）1 [A]01[A]

[A]0L0

L L0LtLL

(A)0

Lt

L0

(5分 （2）ln[A]0xln[A]0

[B]x ln1[B]xln2[B]0 ln ln2[B