JOURNAL OF SCENCE OF HNUE Educanonai Sci.. 2013. Vol. 58. No. 8, pp. 94-102 This paper is available onlme at http://sldb.hnue.edu.vn
SL/ DUNG MOT SO BAI TOAN NHAN T H l / C PHAN HOA HVV CO L 6 P 11 NANG CAO TRONG DAY HOC DAT VA GIAI QUYET VAN ot
Dang Thi Oanh^ Tran Ngoc Huy^
^Khua Hua hpe. TrUdng Dai hoc Supham Hd Npi, -Nhd xudt hdn Gido due Viet Nam Tom lat. Xay dung va sii dung bai loan nhan thilc (BTNT) trong day hoc hoa hpc la mol giai phap quan trong de phat huy linh cUc, chii dong, sang tao ciia hoc sinh la mot trong nhu'ng xu hudng doi mdi phifdng phap day hoc hien nay. Chung toi da nghien ciiu each sii dung BTNT trong day hoc dat va giai quyet van de de phat tnen nang life hoc sinh va da tien hanh thi/c nghiem tai trudng THPT Pham Hong Thai, Ha Noi. Kk qua thu difdc la tich cifc, hoan thanh tot muc tieu dS ra gop phan nang cao chat lUdng day hoc bo mon Hod hoc cl trudng THPT.
Tifkhda- Bai toan nhan thiic, hoa hCfu cd I6p 1!, day hoc dat va giai quyet vku dl.
1. Mot dau
1.1. M o t s o van de ve li l u a n ve e a c h sur d u n g B T N T t r o n g day h o c dat va giai q u y e t v a n d e
Ban chat cua day hoc dat va giai quyet co van dt la giao vien dat ra trUdc HS cac vSn de cua khoa hoc (cac bai loan nhan thu'c) va md ra cho cac em nhiing con difdng giai quy6t cac van de do; Viec diau kien qua irinh tiep thu kiSn thiic ciia HS d day dUdc thitc hien ihco hifdng tao ra mot he thdng nhifng tinh hu6ng co van dc, nhu'ng dieu kien bao dam \'icc gi;ii quycl nhiing tinh hudng do va nhiing chi dan cu the cho HS trong qua trinh giai quyet ciic vin de.
Day hoc dat va giai quyel vSn de co ba dac trifng:
- GV dai ra tnfdc HS mol loat nhffng bai loan nhan thifc cd chiia difng mau IhuSn gma cai da biet va cai chifa biet nhUng chiing dUdc cSu true lai mol each sU pham goi la bai toan neu vSn de - drixtic.
- Hpc sinh li6p nhan mau thuin cua BTNT nhd mau thuSn noi tam cQa minh va difpc dat vao linh hudng co vSn dc, ttic la irang thai co nhu cku ben trong mucSn giai quyet bang difdc bai loan do.
Ngay nhan bai 1.V12/2012 Ngay nhan dang'25/7/2013.
Lien he. DAng Tin Oanh, e-mail [email protected] vn 94
Sif dung mot .so hdi tndn nhdn thitc phdn Hod hdu eo ldp U - Ndng can
- Trong va bang each giai quy6t bai toan do ma HS ITnh hoi difdc mot each tif giac lich cifc ea kien thiic, ca each giai va do do co difdc niem vui sifdng ciia siJ nhan thiic sang tao.
Bai toan nhan thiic difdc coi la hat nhan cua day hoc neu vSn d6 - dritxtic [3], tuy nhien BTNT ciing dUdc sii dung trong cae phifdng phap day hoc khac nhif. cac phifdng phap day hoc trifc quan, cac phUdng phap thifc hanh thuc nghiem, cac phifdng phap dung ldi,... khi BTNT difdc SLt dung trong cac phUdng phap nay thi no se hoal hoa cac phifdng phap dd hay no se lam cho cac phifdng phap do tich cdc hdn, hieu qua hdn [4], 1.2. Su dung BTNT trong day hoc neu van de
a) Sii dung BTNT theo quy trinh
BTNT
(mnu ilniaii khach qiiuiO
- G V l l i t r n i g d d i i - HS nghiCLiciru
T i n h huong CO van de
biOn th.inh VIT clui(|Uiin)
- ( i V g i u | i
- IIS liriiiui BTNT Cureka !
b) Ba miic do ciia day hpc ncu van de;
- GV thuc hien toan bo quy trinh (tu neu, tU giai quytt vin de) - Thuyet trinh orixtic;
- GV va HS ciing tim va giai quyet vin de - Dam thoai drixtic, - HS ti( tim ra va tU giiii quySt van dS - nghien cKu drixtic 1.3. Cac bit6c day HS giai quyet van dc su dung BTNT [4]
Budc 1 Lam cho HS hieu ro vin di (BTNT).
BUdc 2. Xac dinh hudng giai quydt, di(a ra cac gia thuyfit.
Budc 3. Kiem tra sit dung ddn ciia cac gia thuyet va xac nhan 1 gia thuyet diing.
Bitdc 4. Ket luan, chi ra cai mdi.
Budc 5. Kiem tra lai va ling dung vao thuc te.
1.4. Cac each sit dung BTNT trong day lioc - Sii dung BTNT khi nghien cilu bai mdt.
- Sil dung BTN'I' trong luyen tap, ciing co vit nang cao kicn thiic.
- Sii dung BTN T de til hoc d nha.
* V;»' du I. Su dting BTNT khi nghien cdu lidi mdi.
a) Npi dung BTNT ve phdn ling halogen hod ankan.
- Quan sat hinh 5.4 SGK Hodlipc 11 - NC cho biet; C H , tac dung vdi Cl, thu dUdc nhiing san pham nao? Riit ra khai niem ve phan dng the.
- Hoan thanh cac phan ling sau va cho biet san pha'm chinh, phu trong moi phan ling va giai thich?
9S
C H , - C H j - C H , + Br2 5—4
Rut ra kha nang tham gia phan ling the cua H d C bac khac nhau va klia nang phan ling ciia cac halogen.
h) Cdc Inldc day HS gidi BTNT
Budc 1 Lam cho HS hieu ro vin de (BTNT)
- (3 ldp 9 HS da biet CH4 + CI2 chi thu dude 1 san pham la CHuCl nhUng khi quan sal hinh ve 5 4 SGK Hod Iwc i 7 - NC ta lai thiy co 4 san pham the. Mau thuin xuat hien.
- GV cho HS viet san pham 2 phan iing sau va ghi ro s6 ti le % cac chat san pham.
(1.1) 2 — clopropan(57%) 1 — clopropan(43%)
CH;, - CH2 - CH., 4- Bi-j ' " ' " ' ) CH3 - C:HBr - C H , + C H , - CHj - CHaBr + HBi (1.2) 97% 3%
Trong phan tii propan co 6 nguyen til H ciia C bac I, co 2 nguyen tu? H cila C bac II, tai sao lUdng siin phixm the vao H ciia C bac II lai nhieu hdn (spc)? Tai sao clo the ditdc ca H 6 C bac I va II con brom chi the dUdc H cf C bac II? Mau thuin xuit hien.
Budc 2, 3. Xac dinh hudng giai quyet, dda ra cac gia thuyet;
- Xuat phat tif ciu tao CH4, cd 4 lien ket vay clo co the the lin lUdt tifng H khong?
96
Su dmig mdt .so bdi lodn nhdn thifc phdn Hod hifu CO ldp 11 -Ndng cao...
- Mdc do phan iing cua H d C bac khac nhau co khac nhau khong?
- Miic do hoat dong hoa hoc ciia cac halogen (CI2. Bi^) cd anh hudng d6n ti le san pham the nhif the nao?
Bifdc 4. Ket luan, chi ra cai mdi.
- Clo da the lin lUdt titng nguyen tu: H trong CH4.
- H cl C bac cang cao thi kha nang bi th6 cang d l dang: H - C > H - C > H - C - Halogen hoat dong hoa hoc cang yeu thi tinh chon loc cang eao Br2 > CI2.
Bifdc 5. Kiem tra lai va iing dung.
TCf CH:, - CI-I2 - CH-i mudn didu che CH., - CH(OH) - CH3 thi phai theo sd dd:
CH, -~ C H . ^- CH3 - - ^ CH3 - CHBr - CH3 ^^^^> C H , - CH(OH) - CH, Thi du 2. Sii dung BTNT dang dinh lifdng trong luyen tap
a) Noi dung BTNT
Khi cho 27,6 gam ancol ddn chde bang CuO, dun ndng thu dUdc m gam hdn hdp X (con ancol dU). LSy ni gam hdn hdp X cho tac dung vdi Na dd thu difdc 8,96 lit khi H2 d dktc va dung dich Y. Mai khac liy A-ni gam hdn hdp X cho tae dung vdi NaaCO;, dd cung dufdc 8,96 lit khi (dklc) (khong tao ra NaHCO.,). Xac dinh cong thdc phan lii ciia ancol
b) Cac budc day HS giai BTNT Budc 1. Lam eho HS hieu ro van de (BTNT)
- Ta da bi6l khi oxi hoa ancol bang CuO, t^ thu dUdc andehit hoac xeton va nUdc, co the cdn ancol du nhyung cac chit nay dSu khdng phan lifng difdc vdi Na^CO ^ giai phdng khi CO2. Nhung theo gia thiet thi X tac dung vdi Na2C0:, giai phdng khi CO2. Vay X con chda chit nao khac. Mau thuan xuSt hien.
- Khi xae dinh dUdc thanh phin cua X, de tim i\l„ncoi ta phai tim HU,,™;. Mudn tim
• i-ancoi ta phai bat dau tif dau? Mau thuSn xuit hien.
- Ta cd the rut ra he qua hay mot cong thu'c chung nao dd de giai bai nay nhanh khdng?
Budc 2. Xac dinh hudng giai quydt
- Oxi hoa ancol bang CuO. t^ thu dUdc andehit hoac xeton, nUdc, ancol du va cd the cd axit cacboxylic.
- Ta difa vao 2 dii kien so mol H2 va s6 mol khi CO2 de thiet lap mdi lien he vdi sd mol ancol.
- Neu ta thiel lap dUde mdi lien he giiia sd mol H2 va s6 mol khi CO2 de thiel lap mdi lien he vdi s6 mol ancol thi dd chinh la he qua can xac dinh
Budc 4. Ti6n hanh giai - Cac qua trinh oxi hoa ancol:
vr
Dang Thi Oanh, Tran Ngoc Huy
° 2RCH2OH -^^^^^ 2 R C H 0 + H2O
Vay hdn hop X gom; RCHO . r m o l RCOOH ymol; R C H j O H dU 2 moi; HoO I + y mol.
Nen tdng so mol RCH2OH x + >' + z mol.
- Trong dd (RCOOH y mol; RCHjGH dU 2 mol; H j O x A \ mol) Phan ling vdt Na tao ra;
"H, = 5 mol - Vdi NaaCO, chi co RCOOH phan ifng;
2 R C 0 0 H + Na2C03 -^ 2 R C 0 0 N a + CO2 V !//2 mol
- M6i lien he giiia so mol ancol, H2 va CO2 la;
"anmi = 2(nH, " "CO,) = 2(0, 4 - ^ -) = 0. 6 mol
MRCH,OH = ~ = 46 => C2H5OH U, D
Budc 5. Ket luan, chi ra cai mdi va phUdng phap giai ciia dang dang bai toan nay - Oxi hoa ancol bSng CuO thu dUdc andehit hoac ancol va cd the cd axil cacboxylic.
- Xudt phai tif s6 mol Ho va s6 mol khi CO2 ta thiel lap mdi hen he vdi s6 mol ancol do chinh la he qua cdn dimg trong giai loan trac nghiem la:
Thi du 3: Sii dung BTNT ve lien ket hidro de ciing cd va nang cao kien thdc trong viec tu hoc
Ancol etyltc (CoHsOH) va nU6c (H2O) deu co lien ket hidro va
^/CaHsOH = 4(J ^ . 1 / H , 0 = 18
Si'f dimg mot .so hdi todn nhdn thifc phdn Hod hifii cO ldp 11 - Ndng cao...
nhung nhiet do soi ciia C2H5OH la 78.3°C' thSp hdn ciia H2O la 100"*:'?
Budc 1. Lam cho HS hie'u ro van de
Nhiet dp soi ciia cac chit phu thuoc vao phan tit khdi ciia cac chit va lien ket lien phan tif (dac biet la lien ket hidro), ca C2H5OH va H2O deu cd lien ket hidro ma phan tii khoi ciia
-'*/c,H50H=46 > . U H , O = 1 8
nhung tai sao nhicl dp soi cua C2H5OH < H2O. Mau thuin xuat hien.
Budc 2. Xac dinh hUdng giai quyet, de xuil cac gia ihuyet
- Dp ben hen ket hidro giifa cac phan tii C2I-I5OH < giiia cac phan tii H2O.
- Sd lien ket hidro giua cac phan tu C2H5OH < giiia cac phan tii H2O Budc 3. Xac nhan mpt gia thuyet diing
- Giiia cac phan td ancol:
H - O^- . . H^+ - O
I
va giiia cac phan tii nudc;
C2H5 C2H5
• 0
H H
Mac dCi dien tich am tren O ciia ancol > nUdc nhUng dien tich dUdng tren H ciia nudc > ancol nen dp ben lien ket hidro gifla hai loai lien ket tren khong khac nhau nhieu, khong the gay sfl chenh lech nhiet do hai chit ldn nhu vay. (loai gia thiet nay)
- Do Irong nUdc chi co 2 trung tam tao lien ket hidro la hat nguyen tut H. O trong nhdm OH trong khi dd d nildc cd 3 trung tam tao lien ket hidro la 2 nguyen t i H vit 1 nguyen til O nen s6 hen ket hidro giiia cac phan tut nudc nhieu hdn gifla cac phan tii ancol, day chinh la nguyen nhan lam cho H2O du cd phan tfl khdi nho hdn C2H.5OH nhung nhiet dd soi vin cao hdn.
Bfldc 4. Chi ra kien thiic mdi
- Do b6n lien ket va s6 lugng lien ket hidro anh hfldng Idn den nhiet do sdi cua cac chit.
- Do H2O cd lien ket H d 3 trung tam nen d trang thai rin (nude da) thi the' tich ldn hdn the tich long.
Budc .5 Kie'm tra lai kien thflc va van dung giai y b).
2. Noi dung nghien curu 2.1. Thu'c nghiera sU pham
Tien trinh thflc nghiem sfl pham;
Dang Thi Oanh. Trin Ngoc Huy
- Dia diem trudng THPT Pham Hong Thai - Ha Ndi; Giao vien day: Pham Thi Xo.
- Chiing toi da tien hanh day phin hidrocacbon vao hpc ki II cua nam hoc 2010 - 2011. Ci ldp ddi chiing la 11A5 (46 HS) GV sti dung giao an binh thudng, ldp thifc nghiem la 11A4 (46 HS) GV thiet ke giao an cd suT dung BTNT. Ca hai ldp nay din hoc chUdng trinh Hda hpc 11 - nang cao, deu khdng phai la ldp chon, cung GV day.
- Chiing toi tien hanh kiem tra 2 bai kiem tra 45 phdt (trac nghiem va tU luan) d hai ldp (de bai kiem tra nhd nhau, cimg dap an va ciing GV chim) de' danh gia hieu qua ciia
viec sii dung BTNT. «>,, - Ket qua thUc nghiem sU pham diidc trinh bay trong cac Bang 1 va 2 dUdi day-
B a n g / . Phdn phoi tdn so, tdn sudt vd tdn sudt lily tich (Ldp 11A4 vd HAS triidng THPT Pham Hdng Thdi - bdi sd I) Diem
0 1 2 3 4 5 6 7 8 9 10
E
So HS dat diem Xi DC
0 1 1 2 2 10 9 12 6
TN 0 0 0 I 3 6 5 10 11 2 1 8 1 j 2 46 1 46
% HS flat diem Xi DC
0,00 2,17 2,17 4,35 4,35 21,74 19,57 26,09 13,04 4,35 2,17 100,00
TN 0,00 0,00 0,00 2,17 6,52 13,04 10,87 21,74 23,91 17,39 4,35 100,00
% HS d a t diem Xi troi xuong DC
0,00 2,17 4,34 8,69 13,04 34,78 54,35 80,44 93,48 97,83 100,00
TN 0,00 0,00 0,00 2,17 8,69 21,73 32,60 54,43 78,25 95,65 100,00
120
•g too
Z 80 x
.1 60 I' 40
7 20
. - ^
• / V
l^
, . ^ l ^
• DC
2 3 4 5 6 7 8 9 iO Dicm Xi Hinh I. Bd thi ditdng hiy tinh
100
Su ditng mdt so hdi todn nhdn tlufe phdn Hod hUu ca Idp 11 - Ndng cao...
Bdng 2. Phdn phoi tdn so, tdn sudt vd tdn sudt luy tich (Ldp 11A4 vd HAS trudng THPT Pham Hong Thdi - bdi sd 2) Diem
0 1 2 3 4 5 6 7 8 9 10
E
So HS dat diem Xi DC
0 1 I 3 5 9 10
6 5 5 1 46
TN 0 0 I
% HS dat diem Xi DC
0,00 2,17 2,17 I 6,52 2
4 12 10 7 6 3 46
10,87 19,57 21,74 13,04
TN 0,00 0,00 2,17 2,17 4,53 8,70 26,09 21,74 10,87 15,22 10,87
2,17 100,00
13,04 6,52 100,00
% HS dat diem Xi trd xuong DC
0,00 2,17 4,34 10,86 21,73 41,3 63,04 76,08 86,95 97,82 100,00
TN 0,00 0,00 2,17 4,34 8,69 17,39 43,48 65,22 80,44 93,48 100,00
2 3 4 5 6 7 8 Hinh 2. Do thi dtldng biy tich
9 10 DiemXi
Phan tich ket qua thflc nghiem;
- Tren cd sd ket qua thflc nghiem cho thiy, chit Ifldng hoc tap cua hoc sinh cac ldp thflc nghiem cao hdn cac ldp doi chflng, the' hien;
+ Die'm trung binh cac bai kie'm tra cua ldp TN luon cao hon cac Idp DC.
+ Ti le % hoc sinh dat die'm ySu kem, trung binh cua cac Idp DC cao hdn ldp TN, con ti le % hoc sinh dat die'm kha, gioi cua ldp TN luon cao hdn cua cac ldp DC.
+ D 6 thi cac dudng luy tich cua cac ldp TN luon nam ve ben phai va d phia dfldi dd thi cac dfldng luy tich cua cac Idp DC (Hinh 1 va Hinh 2), dieu dd chflng td chit Ifldng
Dang Thi Oanh, Trin Ngoc Huy
nam vflng kien thiic cua hoc sinh cac ldp TN tdt hdn, ddng deu hdn so vdi cac ldp DC.
- Ngoai ket qua thuc nghiem tif diem so bai kiem tra, danh gia, chiing toi con cd sif so sanh ve tinh thSn thai do hpc tap, khong khi gid hoc thi ldp thdc nghiem ludn cho ket qua tich cue hdn ldp doi chiing.
- Nhu vay, viec sd dung BTNT trong day hpc dat va giai quySt vSn de giiip HS chu dong tiep thu kien thiic va phat trien tu duy sang tao ciia HS.
3. Ket luan
Tren day chiing toi da tien hanh sd dung cac BTNT trong day hoc dat va giai quySt vin de de day hoc bai mdi, de dn tap cung co, de tu hpc nang cao kiln thiic phin hdp chk hifu cd trong SGK Hod hpc 11 - NC. Chung toi tilp tuc nghien ciiu sif dung cac BTNT vao mot so PPDH khac de phat huy tinh tich cue, chu ddng, phat huy nang lUc hoc sinh vao s6 bao sau. Chung tdi da chiing minh rang, khi xay ddng dUdc he thdng BTNT dap Ung tdt nhung yeu cau dat ra va van dung nd mdt each sang tao trong day hpc se mang lai hieu qua nhu muc tieu dd ra, dieu do se tao ra sU thay doi ve chSt trong each day - hoc Hoa hoc, gdp phin doi mdi phUdng phap day hpc d nUdc ta hien nay.
TAI LIEU THAM KHAO
[1] Nguyin Ngoc Quang, 1994. Li ludn day hpc Hod hoc, tap 1. Nxb Giao due.
[2] Nguyin CUdng, 2007. Phuong phdp day hpc Hod hpe d trudng pho thdng vd dai hpc. Nxb Giao due.
[3] Nguyin Huu Dinh (nhieu tac gia), 2007. Hod bpc 11 - Ndng cao. Nxb Giao due Viet Nam
[4] Trin Qudc Sdn, 2008. Gido trinh co .sd hod hpc hixu co, tap 1. Nxb Dai hpc SU pham.
[5] Prof. Bernd Meier, 2009. Li luan day hue hien dai. Potsdam, Ha Npi.
ABSTRACT
Using recognition problems (RP) in 11"' Grade organic chemistry to teach recognition and solution of problems
Designing and using recognition problems (RP) in the teaching of chemistry is a way lo encourage greater activity, self-reliance and creativeness. This is an innovative trend in teaching and learning methods today [1]. We've used RP to develop students' ability to recognize and solve problems. This has been tested at Pham Hong Thai High School in Hanoi and the results were very good, meeting the objectives and improving the quality of chemistry teaching in this secondary school.
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