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KHOA GIÁO DỤC ĐẠI CƯƠNG THI HỌC KÌ I (LẦN 2) - HỆ CAO ĐẲNG BỘ MÔN: VH - NN MÔN: HÓA HỌC ĐẠI CƯƠNG Ngày thi:………………............... THỜI GIAN LÀM BÀI: 60 phút Họ tên sinh viên:…………………………………… MSSV: ……………………………………………… Lớp: ……………………………………………........ Giáo viên coi thi 1 Giáo viên coi thi 2 Giáo viên chấm thi 1 Giáo viên chấm thi 2 Điểm SV chỉ được dùng bảng tuần hoàn để tra cứu. PHN 1 : TRC NGHIM T LUN (5 đim) SV trình bày ngắn gọn cách làm ghi kt qu vo cui câu. Câu 1: Cho các phương trình nhiệt hóa học sau: S (thoi) + O 2(k) SO 2(k) H 0 298 = -296,06 kJ S (đơn tà) + O 2(k) → SO 2 (k) H 0 298 = –296,36 kJ Tính biến thiên entanpi tiêu chuẩn của phản ứng: S (thoi) S (đơn tà) ...................................................................... ...................................................................... ...................................................................... ...................................................................... ...................................................................... ...................................................................... ...................................................................... ...................................................................... ...................................................................... .................................................. ................... Trang 1/7 - Mã đề thi 169

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KHOA GIO DC I CNG

THI HC K I (LN 2) - H CAO NG

B MN: VH - NN

MN: HA HC I CNG

Ngy thi:...............

THI GIAN LM BI: 60 pht

H tn sinh vin:

MSSV:

Lp: ........Gio vin coi thi 1

Gio vin coi thi 2

Gio vin chm thi 1

Gio vin chm thi 2im

SV ch c dng bng tun hon tra cu. PHN 1: TRC NGHIM T LUN (5 im)SV trnh by ngn gn cch lm v ghi kt qua vao cui cu.

Cu 1: Cho cc phng trnh nhit ha hc sau: S(thoi) + O2(k) SO2(k) (H0298 = -296,06 kJ

S(n t) + O2(k) SO2 (k) (H0298 = 296,36 kJ

Tnh bin thin entanpi tiu chun ca phn ng: S(thoi) ( S(n t) ........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................ ..................................................................................................................................................................................................................................................................................................................................................

H0 = 0,3 kJCu 2: Ha tan 100g CuSO4.5H2O vo 400g dung dch CuSO4 4%. Tnh nng phn trm ca dung dch thu c........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................ ..................................................................................................................................................................................................................................................................................................................................................

C% = 16%Cu 3: Phn ng phn hy phng x ca mt ng v l bc 1 v c chu k bn hy t1/2 = 15 pht. Sau bao lu 80% ng v b phn hy?........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................ ..................................................................................................................................................................................................................................................................................................................................................

t = 34,83 phtCu 4: Cho phn ng : 2N2O5 ( 4NO2 + O2. 250C c hng s tc l k1 = 1,72.10-5 s-1.

350C c hng s tc l k2 = 6,65.10-5 s-1. Tnh nng lng hot ha Ea ca phn ng.(Cho R = 8,314 J/mol.K = 1,987 cal/mol.K)........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................ ..................................................................................................................................................................................................................................................................................................................................................

Ea = 103 kJCu 5: 1500C mt phn ng kt thc trong 16 pht. Tnh thi gian phn ng kt thc nhit 2000C. Cho bit h s nhit ca phn ng ny bng 2,5......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................................................................................

t = 0,164 phtPHN 2: TRC NGHIM KHACH QUANA. BANG TRA LI

Hng dn lm bi :

Chn : b : chn li :

Cu1234567891011121314151617181920

A

B

C

D

B. CU HOI Cu 1: Tnh (H0298 ca phn ng: 2Mg(r) + CO2(k) 2MgO(r) + C(gr)Bit rng (H0298,s (CO2) = 393,5 kJ

(H0298,s (MgO) = 601,8 kJ

A. +208,3 kJB. 208,3 kJC. +810,1 kJD. 810,1 kJCu 2: S proton, ntron, electron ca nguyn t ln lt l :

A. 25; 55 v 25.B. 25; 25 v 30.C. 25; 30 v 25.D. 55; 25 v 25.Cu 3: Nguyn t c Z = 26 thuc loi nguyn t no ?

A. Nguyn t s.B. Nguyn t p.C. Nguyn t d.D. Nguyn t f.Cu 4: Phn t c momen lng cc bng khng ( = 0) l

A. cacbon tetraclorua.B. Natri clorua.C. nc.D. amoniac.Cu 5: Electron cui cng ca nguyn t nguyn t X ng vi: n = 3, = 2, m = 0, ms = -1/2 .Vy trong bng tun hon, X thuc s

A. 25.B. 28.C. 26.D. 27.

Cu 6: B cc s lng t ph hp l :

A. n = 1, l = 0, ml = 1.B. n = 3, l = 0, ml = 2.

C. n = 3, l = 3, ml = 0.D. n = 4, l = 3, ml = 1.Cu 7: 1000oC hng s cn bng ca phn ng: FeO(r) + CO(k) (Fe(r) + CO2(k) bng 0,5; nng ban u ca cc cht nh sau: [CO] = 0,05M; [CO2] = 0,01M.

Nng cc cht lc cn bng l :

A. [CO] = 0,02M; [CO2] = 0,04MB. [CO] = 0,04M; [CO2] = 0,02MC. [CO] = 0,02M; [CO2] = 0,01MD. [CO] = 0,01M; [CO2] = 0,02MCu 8: Cation X+ c phn lp cui l 2p6. Vy trong bng tun hon, X thuc th :

A. 12B. 11C. 13D. 10Cu 9: Entropi (S) ca cht no di y ln nht 25oC v 1atm?

A. C2H6 (k)

B. CH4 (k)

C. C3H6 (k)

D. C4H10 (k)Cu 10: Cho phn ng: Mg + HNO3 ( Mg(NO3)2 + N2O + H2O

Tng cc h s nguyn, ti gin ca phn ng l:

A. 20.B. 24.C. 22.D. 26.Cu 11: Cho s pin : () Zn | Zn2+ || 2I | I2(k) | Pt(r) (+). Bit th kh tiu chun ca Zn2+/Zn l

-0,763V v ca I2 /I- l 0,540V th sc in ng chun ca pin bng:

A. +1,303V.B. 0,223V.C. +0,223V.D. 1,303V.Cu 12: Trong phn t BF3, ln ca gc lin kt F-B-F l

A. 900.B. 1800.C. 1200.D. 109,50.Cu 13: Nhng c im ph hp vi phn t NH3 l

A. Cu trc thp, phn cc.B. Cu trc t din, khng phn cc.

C. Cu trc tam gic phng, gc ha tr 1200.D. Cu trc t din, gc ha tr 1070.Cu 14: Phng trnh c bn ca nhit ng ha hc l

A. G = H + TS.B. G = H TS.C. G = H + TS.D. G = H TS.Cu 15: Bit rng: khi nhit tng ln 100C, tc ca mt phn ng tng 2 ln. Nu phn ng ang xy ra 20oC, tng tc phn ng ln 32 ln th phi thc hin phn ng

A. 120oC.B. 70oC.C. 60oC.D. 80oC.Cu 16: Cho bit m in cc nguyn t Al (1,61); S (2,58). Nhm sunfua cA. lin kt ion.B. lin kt cng ho tr phn cc.C. lin kt cng ho tr khng phn cc.D. lin kt kim loi.Cu 17: Theo thuyt lin kt ha tr VB (Valence Bond), lin kt trong phn t HF c hnh thnh do s xen ph gia cc orbital

A. 2p (ca F) v 1s (ca H).B. 2s (ca F) v 1s (ca H).C. 1s (ca F) v 1s (ca H).D. 2p (ca F) v 2s (ca H).Cu 18: Lin kt N H trong phn t NH3 l lin kt

A. ion.

B. cng ha tr phn cc.C. cng ha tr khng phn cc D. phi tr.

Cu 19: Cho phn ng n gin : 2A (k)+ B (k) ( C (k), c hng s vn tc k = 0,5 .

Nng ban u ca A l 6M, ca B l 5M. Vn tc phn ng khi c 55% cht B tham gia phn ng c gi tr l

A. 0,5625.B. 0,3500.C. 2,2500.D. 0,2813 .Cu 20: Cho phng trnh nhit ha hc sau :

2H2(k) + O2(k) 2H2O(l) ; Ho298 = -571,68 kJ

Nhit to thnh ca H2O(l) l

A. 571,68 kJ/mol.

B. +571,68 kJ/mol.

C. 285,84kJ/mol.

D. + 285,84kJ/mol

Ngay 4 thang 3 nm 2015Khoa/b mn Ging vin ra

Th Ngc Mai

Mn

55

25

Trang 4/4 - M thi 169

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