. 49912009 49912014 H2O2 Ex H2O2()+Br2() Ex H2O2()+KI()

/

r aA + bB cC + dD

R = k [] n k n

H2O2

( n=1 )O2H2O2

O2100ml510ml(100ml)250ml100mlH2O2KI35

0.05N & 0.1N KI10ml0.05N0.083g 10ml0.1N0.166g 10ml(166.01g/mol)3%5ml0.86ml 35% H2O2 10ml

3. !4.5.5ml 3% H2O26.10ml 0.05N KI7.(0.05N KI)30O2(15~30min)8. 45 O2V 9. 0.1N KI step4.-8.10. O2H2O2

t =k ()

t =

0.05M KIV(ml)93Vt(ml)7.514.023.430.035.541.046.051.055.560.5V- Vt (ml)85.579.069.663.057.552.047.042.037.532.5t (s)306090120150180210240270300Log(V- Vt)1.9311.8971.8431.7991.7601.7161.6721.6231.5741.5120.1M KIV(ml)55Vt(ml)8.415.120.226.130.534.037.441.543.545.0V- Vt (ml)46.639.934.828.924.521.017.613.511.510.0t (s)306090120150180210240270300Log(V- Vt)1.691.601.541.461.391.321.251.141.061.00= -0.0015 = -k/2.303k = 0.0015*2.303 = 0.0035= -0.0026 = -k/2.303k = 0.0026*2.303 = 0.0060 http://case.ntu.edu.tw/hs/wordpress/wp-content/uploads/2010/12/form110.jpghttp://blog.ncue.edu.tw/yangsp/doc/29402