Tiêu đề রাসায়নিক পরিবর্তন MCQ Practice-Sheet_With-Solve.pdf ✅

ivmvqwbK cwieZ©b  Varsity Practice Sheet 1 PZz_© Aa ̈vq ivmvqwbK cwieZ©b Chemical Changes ACS Chemistry Department Gi g‡bvbxZ eûwbe©vPwb cÖkœmg~n ivmvqwbK wewμqv I wMÖb †Kwgw÷a 1. wMÖb †Kwgw÷ai eviwU bxwZi g‡a ̈ †KvbwU mwVK bq? [JU-A 19-20] Kg KuvPvgv‡ji e ̈envi wbivc` `aveK e ̈envi h_vmg‡q `~lY wbqš¿Y cÖfveb cÖ‡qvM DËi: Kg KuvPvgv‡ji e ̈envi e ̈vL ̈v: wMÖb †Kwgw÷ai eviwU bxwZi g‡a ̈ bevqb‡hvM ̈ KuvPvgv‡ji e ̈envi Av‡Q, Kg KuvPvgv‡ji e ̈envi †bB| wMÖb †Kwgw÷ai 12wU bxwZ: 1| eR© ̈ c`v_© †ivaKiY, 2| m‡e©vËg GUg B‡Kv‡bvwg, 3| b~ ̈bZg SzuwKi c`v‡_©i e ̈envi 4| wbivc` †KwgK ̈vj cwiKíbv, 5| wbivc` `aveK e ̈envi, 6| wewμqvi kw3 `ÿZv cwiKíbv, 7| bevqb‡hvM ̈ KuvPvgvj e ̈envi, 8| b~ ̈bZg DcRvZK, 9| cÖfveb cÖ‡qvM, 10| cÖvK...wZK iƒcvšÍi cwiKíbv, 11| h_vmg‡q `~lY wbqš¿Y, 12| `yN©Ubv cÖwZ‡iva| 2. wMÖb †Kwgw÷ai eviwU bxwZi g‡a ̈ wb‡Pi †KvbwU mwVK bq? [JU-A 19-20] wbivc` `aveK e ̈envi eR© ̈ c`v_© †ivaKiY b~ ̈bZg DcRvZK cÖvK...wZK †KwgK ̈vj cwiKíbv DËi: cÖvK...wZK †KwgK ̈vj cwiKíbv e ̈vL ̈v: wMÖb †Kwgw÷ai eviwU bxwZi g‡a ̈ wbivc` †KwgK ̈vj cwiKíbv Av‡Q| cÖvK...wZK †KwgK ̈vj cwiKíbv †bB| 3. †KvbwU mwVK bq? [JU-D 18-19, PSTU 18-19] GUg B‡Kv‡bvwg = (Kvw•ÿZ Drcv‡`i fi/†gvU Drcv‡`i fi)% B-d ̈v±i = e‡R© ̈i †gvU fi/ Drcv‡`i †gvU fi GUg B‡Kv‡bvwgi gvb hZ D”P GKwU Drcv`b cÖwμqv ZZ meyR B-d ̈v±i hZ †ewk Zv ZZ cwi‡ek evÜe DËi: B-d ̈v±i hZ †ewk Zv ZZ cwi‡ek evÜe 4. †KvbwU wMÖb †Kwgw÷ai 12wU bxwZi g‡a ̈ bq? [CU 16-17] b~ ̈bZg DcRvZK bevqb‡hvM ̈ KuvPvgvj e ̈envi m ̄Ív †KwgK ̈vj e ̈envi kw3 `ÿZv cwiKíbv DËi: m ̄Ív †KwgK ̈vj e ̈envi 5. wMÖb †Kwgw÷ai 12wU bxwZi g‡a ̈ mwVK bqÑ [IU-D 19-20] eR© ̈ c`v_© †ivaKiY m‡e©vËg GUg B‡Kvbwg kÖwg‡Ki ̄^v ̄’ ̈ †mev `yN©Ubv cÖwZ‡iva DËi: kÖwg‡Ki ̄^v ̄’ ̈ †mev 6. †Kv‡bv ivmvqwbK wewμqv cwi‡ek mnvqK wKbv Zv eySv hvqÑ [CoU-B 15-16] R¡vjvwbi e ̈envi wbqš¿Y Øviv weï× wewμqK Øviv Atom Economy Gi D”Pgvb Øviv Yield Gi D”PgvÎv Øviv DËi: Atom Economy Gi D”Pgvb Øviv e ̈vL ̈v: kZKiv GUg B‡Kv‡bvwg Øviv †Kv‡bv c×wZi mdjZv Ges ivmvqwbK wewμqv cwi‡ek mnvqK wKbv Zv eySv hvq| 7. wMÖb nvDR cÖwZwμqvi KviY Kx? [Agri.cluster 20-21] meyR DwTM¢` Bbd«v‡iW iwk¥ UV iwk¥ iÄb iwk¥ DËi: Bbd«v‡iW iwk¥ e ̈vL ̈v: Av‡jvKiwk¥ f~-c„‡ô cwZZ n‡q †ewk Zi1⁄2‣`‡N© ̈ IR (Infrared) iwk¥ •Zwi K‡i, hv KuvP †f` Ki‡Z cv‡ibv Ges djkÖæwZ‡Z ZvcgvÎv evovq| 8. †KvbwU wMÖb `aveK? [MBSTU-C 18-19] KwVb CO2 CCl4 CHCl3 mycviwμwUK ̈vj CO2 DËi: mycviwμwUK ̈vj CO2 9. ETP wK? [SGVC 18-19] Effluent treatment plant Efficient transportation Efficient transfer process Effluent treated product DËi: Effluent treatment plant e ̈vL ̈v: ivmvqwbK wkí KviLvbvi eR© ̈ cvwb ev Zij c`v‡_© •Re I A‣Re c`v_© wgwkÖZ _v‡K| G eR© ̈ cvwb‡K Effluent ejv nq| Giƒc wkí KviLvbvi Effluent †_‡K ÿwZKi ivmvqwbK c`v_©‡K c„_K Kivi cÖwμqv‡K Effluent treatment plant ev ETP e‡j| 10. FeO + CO  Fe (Kvw•ÿZ Drcv`) + CO2| wewμqvwUi GUg B‡Kv‡bvwg KZ? 55.93% 65.39% 55.39% 56.39% DËi: 55.93% e ̈vL ̈v: GUg B‡Kv‡bvwg = 55.85 55.85 + 44 × 100% = 55.93% 11. C2H4 + 1 2 O2 Ag- catalyst  250C C2H4O| wewμqvwUi %AE = ? 100% 0% 50% 90% DËi: 100%
2  Chemistry 1st Paper Chapter-4 12. KI + AgNO3  AgI (Kvw•ÿZ Drcv`) + KNO3| D3 wewμqvq 10 kg AgI Drcbœ Ki‡Z 3 kg KNO3 Drcbœ nq| B– d ̈v±‡ii gvb = ? 0.4 0.3 0.5 0.2 DËi: 0.3 e ̈vL ̈v: B-d ̈v±i = cÖwμqvq †gvU e‡R© ̈i fi (kg) Drcv‡`i †gvU fi (kg) = 3 10 = 0.3 13. CH2 = CH – CH2Cl + H2O → CH2 = CH2 – CH2OH + HCl wewμqvq CH2 = CH2 – CH2OH Drcv` Ges HCl eR ̈©| wewμqvwU‡Z E factor KZ? [SUST- A 16-17] 0.36 0.58 0.63 0.72 0.85 DËi: 0.63 e ̈vL ̈v: E factor = 36.5 12 3 + 6 1 + 16  1 = 0.63 wewμqvi w`K, Zv‡cvrcv`x I Zvcnvix wewμqv 14. †Kvb cwieZ©bwU Zv‡cvrcv`x? [RU-C 20-21] H2O(l)  H2O(g) H2O(g)  H2O(l) H2O(s)  H2O(l) H2O(s)  H2O(g) DËi: H2O(g)  H2O(l) e ̈vL ̈v: Zvc‡kvlY ZvceR©b KwVb Zij ev®ú Zvc‡kvlY ZvceR©b H2O(l)  H2O(g); Zvc‡kvlY H2O(g)  H2O(l); ZvceR©b H2O(s)  H2O(l); Zvc‡kvlY H2O(s)  H2O(g); Zvc‡kvlY 15. †KvbwU Zvcnvix wewμqv? [KU-A 19-20] CH4 + 2O2 = CO2 + 2H2O C + O2 = CO2 2H2 + O2 = 2H2O N2 + O2 = 2NO DËi: N2 + O2 = 2NO 16. wb‡Pi †KvbwU Zv‡cvrcv`x wewμqv msNU‡bi kZ©? [BSMRSTU-C 19-20] eÜb •Zwi‡Z wbM©Z kw3 < eÜb fvO‡Z †kvwlZ kw3 wewμq‡Ki Af ̈šÍixY kw3 > Drcv‡`i Af ̈šÍixY kw3 wewμq‡Ki Af ̈šÍixY kw3 < Drcv‡`i Af ̈šÍixY kw3 eÜb •Zwi‡Z wbM©Z kw3 > eÜb fvO‡Z †kvwlZ kw3 DËi: eÜb •Zwi‡Z wbM©Z kw3 > eÜb fvO‡Z †kvwlZ kw3 e ̈vL ̈v: Drcv‡`i eÜb MVb Ki‡Z wegy3 kw3 hLb wewμq‡Ki eÜb fvO‡Z cÖ‡qvRbxq kw3i †P‡q †ewk nq, ZLb wewμqv †K›`a n‡Z Zvc wbM©Z nq I †mwU Zv‡cvrcv`x nq| 17. †Kvb wewμqvwU Zv‡cvrcv`x? [SAU 14-15] 2NH3(g)  N2(g) + 3H2(g) N2(g) + O2(g)  2NO(g) N2(g) + 3H2(g)  2NH3(g) 2SO2(g)  2SO3(g) + O2(g) DËi: N2(g) + 3H2(g)  2NH3(g) e ̈vL ̈v: †Kv‡bv wewμqvi H = (+)ve n‡j wewμqvwU Zvcnvix Ges H = (–)ve n‡j wewμqvwU Zv‡cvrcv`x| N2(g) + 3H2(g)  2NH3(g); H = – 92.38 kJ/mol Shortcut: Zv‡cvrcv`x wewμqvi D`vniY- nvB Kvjv mv_xi wgZv bvB H2 + O2  H2O C + O2  CO2 SO2 + O2  SO3 CH4 + O2  CO2+H2O N2 + H2  NH3 18. DfgyLx wewμqv‡K GKgyLx Kivi †ÿ‡Î †KvbwU cÖ‡hvR ̈? [MBBS 02-03] wewμqK ̧‡jv hw` KwVb ev Zij nq Ges GKwU Drcv` M ̈vmxq nq GKwU Drcv` `aeY †_‡K Aatwÿß n‡j †Kv‡bv Drcv`‡K ivmvqvwbK wewμqvi gva ̈‡g mwi‡q wb‡j me ̧‡jv DËi: me ̧‡jv e ̈vL ̈v: DfgyLx wewμqv‡K GKgyLx Kivi †ÿ‡Î wb‡¤œv3 kZ© ̧‡jv cÖ‡hvR ̈: i. wewμqK ̧‡jv hw` KwVb ev Zij nq Ges GKwU Drcv` M ̈vmxq nq| †hgb- Zn + H2SO4  ZnSO4 + H2(g) ii. GKwU Drcv` `aeY †_‡K Aatwÿß n‡j| †hgb- NaCl + AgNO3  NaNO3 + AgCl(s) iii. †Kv‡bv Drcv`‡K ivmvqwbK wewμqvi gva ̈‡g mwi‡q wb‡j| 19. †KvbwU DfgyLx wewμqvi D`vniY bq? [MBBS 01-02] CuSO4.5H2O(s) ⇌ CuSO4(s) + 5H2O(g) NH4Cl(s) ⇌ NH3(g) + HCl(g) CH3COOC2H5(l) + H2O(l) ⇌ C2H5OH(l) + CH3COOH(l) 2KClO3(s) ⇌ 2KCl(s) + 3O2(g) DËi: 2KClO3(s) ⇌ 2KCl(s) + 3O2(g) e ̈vL ̈v: 2KClO3(s)  2KCl(s) + 3O2(g) GKwU GKgyLx wewμqv| KviY KCl I O2 KL‡bvB KClO3 •Zwi K‡i bv| 20. GKwU Ave× cv‡Î nvB‡Wav‡Rb I Mvp †e ̧wb e‡Y©i Av‡qvwWb wb‡¤œi KZ ZvcgvÎvq (C) †i‡L w`‡j nvB‡Wav‡Rb Av‡qvWvBW Drcbœ nq? [BDS 10-11] 450 550 350 250 DËi: 450 e ̈vL ̈v: H2(g) + I2(g) 450C 2HI(g)
ivmvqwbK cwieZ©b  Varsity Practice Sheet 3 wewμqvi nvi 21. aA  bB wewμqvwUi †ÿ‡Î †KvbwU wewμqvi nvi wb‡`©k K‡i? [DU 20-21] – d[A] dt – 1 a d[A] dt – d[B] dt – 1 b d[A] dt DËi: – 1 a d[A] dt e ̈vL ̈v: aA  bB wewμqvwUi nvi (r) = – 1 a d[A] dt = + 1 b d[B] dt 22. Zv‡cvrcv`x wewμqvq ZvcgvÎv evov‡j †KvbwU mZ ̈ bq? [DU-A 17-18] wewμqv nvi K‡g mvg ̈ve ̄’v ev‡g hvq wewμqv nvi ev‡o mwμqY kw3 aaæe _v‡K DËi: wewμqv nvi K‡g e ̈vL ̈v: Zv‡cvrcv`x wewμqvi †ÿ‡Î ZvcgvÎv evov‡j mvg ̈ve ̈ ̄’v ev‡g m‡i Av‡m wKš‘ wewμqvi nvi K‡g bv| ZvcgvÎv evov‡j wewμqvi nvi evo‡e| KviY-(i) msNl© evo‡e (ii) mwμq AYyi msL ̈v evo‡e (iii) k = Ae – Ea RT 23. †KvbwU A + 2B  P wewμqvwUi mwVK nvi-mgxKiY wb‡`©k K‡i? [DU 16-17] – d[A] dt = k [A] [B] – d[B] dt = k [A] [B]2 d[P] dt = k [P] d[P] dt = k [A] [B] DËi: d[P] dt = k [A] [B] e ̈vL ̈v: †Kv‡bv ivmvqwbK wewμqvi nvi wewμq‡Ki mwμq f‡ii mgvbycvwZK| Avi cÖwZwU wewμq‡Ki †h mnM _vK‡e Zv m~P‡K cwiYZ n‡e| A + 2B  P wewμqvi nvi = – d[A] dt = – 1 2 [dB] dt = d[P] dt = k [A] [B]2  = 1 I  = 2 n‡j, wewμqvi nvi = – d[A] dt = – 1 2 [dB] dt = d[P] dt = k [A] [B] 24. A + 2B  D wewμqvi †ÿ‡Î wewμqvi nvi mgxKiY n‡jv, rate = k [A] [B]| hw` Dfq wewμq‡Ki NbgvÎv wØ ̧Y Kiv nq, Zvn‡j wewμqvi nvi e„w× cv‡e- [DU 14-15] 2 times 4 times 6 times 8 times DËi: 4 times e ̈vL ̈v: wewμqvi nvi = k [A] [B] = k (2 × 2) = 4k A_©vr 4 ̧Y e„w× cv‡e| 25. 2A + 2B  2C + 2D wewμqvwUi ZvrÿwYK MwZi Rb ̈ mwVK ivwkgvjvwU wPwýZ Ki| [DU 04-05] + d[A] dt – d[A] dt + 1 2 d[D] dt + d[C] dt DËi: + 1 2 d[D] dt e ̈vL ̈v: 2A + 2B  2C + 2D mgxKi‡Yi Rb ̈ v = + 1 2 d[C] dt = + 1 2 d[D] dt = – 1 2 d[A] dt = – 1 2 d[B] dt 26. A + B  C wewμqvwUi MwZ mgxKiY n‡jv V = k [A]2 . A Gi cÖviw¤¢K NbgvÎv wØ ̧Y Kiv n‡j wewμqvwUi cÖviw¤¢K MwZ KZ ̧Y e„w× cv‡e? [DU 04-05, 03-04; JnU 06-07] 2 3 4 8 DËi: 4 e ̈vL ̈v: V = k [A]2 = k(2)2 = 4k A_©vr 4 ̧Y e„w× cv‡e| 27. 2NO(g) + O2(g)  2NO2g) wewμqvwUi cixÿv K‡i wbYx©Z MwZ mgxKiY, v = k[NO]2 [O2]; [NO] Gi cwigvY Av‡Mi †P‡q wØ ̧Y Kiv n‡j Avw` †e‡Mi wK cwieZ©b n‡e? [DU 02-03] AcwiewZ©Z _vK‡e wØ ̧Y n‡e wZb ̧Y n‡e Pvi ̧Y n‡e DËi: Pvi ̧Y n‡e e ̈vL ̈v: v = k [NO]2 = k(2)2 = 4k A_©vr 4 ̧Y e„w× cv‡e| 28. wb‡¤œi †KvbwU nvi aaæeK welqK Avi‡nwbqvm mgxKiY? [BUET 10-11; DU-HEC 21-22;IU 05-06] k = Ne Ea RT k = Ne–N/R k = NeE/T k = Ne – Ea RT DËi: k = Ne – Ea RT 29. cÖwZ 10C ZvcgvÎv e„w×i Rb ̈ wewμqvi nvi KZ ̧Y e„w× cvq? [MBBS 08-09] 3-4 ̧Y 4-5 ̧Y 1-2 ̧Y 2-3 ̧Y DËi: 2-3 ̧Y e ̈vL ̈v: 1889 mv‡j cÖL ̈vZ weÁvbx Gm. Avi‡nwbqvm (S.Arrhenius) me©cÖ_g wewμqvi nv‡ii Ici ZvcgvÎvi cÖfve Abyaveb K‡ib| wZwb cÖgvY K‡ib †h, mvaviYfv‡e 10C ZvcgvÎv e„w×i Rb ̈ cÖvq me wewμqvi nvi wØ ̧Y ev wZb ̧Y e„w× cvq| 30. GKwU wewμqvi ZvcgvÎv 20C e„w× Ki‡j G wewμqvi nvi c~‡e©i Zzjbvq- [SAU 17-18] wØ ̧Y n‡e Pvi ̧Y n‡e wek ̧Y n‡e †Kvb cwieZ©b n‡e bv DËi: Pvi ̧Y n‡e e ̈vL ̈v: cÖwZ 10C ZvcgvÎv e„wׇZ wewμqvi nvi cÖvq 2 ̧Y e„w× cvq|
4  Chemistry 1st Paper Chapter-4 31. wewμqv nv‡ii GKK †KvbwU? [RU-C 21-22; CU-A 20-21, 19-20] †Kvb GKK bvB mol L –1 mol L –1 s –1 ms–1 DËi: mol L –1 s –1 e ̈vL ̈v: wewμqvi nv‡ii GKK = NbgvÎv mgq = mol/L s = mol L –1 s –1 NbgvÎv‡K mol/dm3 GK‡K cÖKvk Ki‡j wewμqv nv‡ii GKK n‡e mol dm–3 s –1 32. cÖwZ GKK mg‡q †Kv‡bv ivmvqwbK wewμqv hZUzKz m¤úbœ nq, Zv‡K H wewμqvi- [IU-D 19-20] nvi e‡j aaæeK e‡j nvi- aaæeK e‡j †Kv‡bvwUB bq DËi: nvi e‡j 33. N2O4 ⇌ 2NO2 Gi †ÿ‡Î wewμqvi nvi †KvbwU? [BSMRSTU-A 19-20] – d(N2O4) dt + d(N2O4) dt + d(NO2) dt – 1 2 d(NO2) dt DËi: – d(N2O4) dt 34. 2A + B  C wewμqvq C MV‡bi nvi 2.2  10–3 mol L–1 min-1 n‡j – d[A] dt Gi gvb KZ? [KUET 18-19] 2.2  10–3 1.0  10–3 1.1  10–3 4.4  10–3 2.4  10–4 DËi: 4.4  10–3 e ̈vL ̈v: Avgiv Rvwb, wewμqvi nvi: – 1 2 d[A] dt = – d[B] dt = d[C] dt †`Iqv Av‡Q, d[C] dt = 2.2  10–3 mol L–1 min–1 GLb, – 1 2 d[A] dt = 2.2  10–3  – d[A] dt = 4.4  10–3 mol L–1 min–1 35. 1 2 A = 2B wewμqvwUi Rb ̈ “A” Gi wbt‡kwlZ n‡q hvIqvi nv‡ii mv‡_ “B” †e‡o hvIqvi nv‡ii m¤úK©Ñ [KUET 17-18] – d[A] dt = 4 d[B] dt – d[A] dt = 1 2 d[B] dt – d[A] dt = 1 4 d[B] dt – d[A] dt = d[B] dt – d[A] dt = 2 d[B] dt DËi: – d[A] dt = 1 4 d[B] dt e ̈vL ̈v: aA ⇋ bB cÖ`Ë wewμqvi nvi = – 1 a d[A] dt = 1 b d[B] dt  – d[A] dt = a b d[B] dt 1 2 A = 2B wewμqvi Rb ̈ a = 1 2 Ges b = 2  – d[A] dt = 1 2 2 d[B] dt = 1 4 d[B] dt 36. A Ges B `yÕwU wewμq‡Ki cÖ‡Z ̈KwUi cÖviw¤¢K NbgvÎv 0.20 mol/dm3 | wewμqvwUi cÖviw¤¢K MwZi nvi 1.6  10–4 mol dm–3 s –1 | wewμqvwUi MwZi nvi aaæeK wbY©q Ki| [KUET 15-16] 4.0  10–3 dm3 mol–1 s –1 8.0  10–6 dm3 mol–1 s –1 4.0  10–3 dm6 mol–2 s –1 8.0  10–6 dm3 mol–2 s –1 1.6  10–4 dm3 mol–1 s –1 DËi: 4.0  10–3 dm3 mol–1 s –1 e ̈vL ̈v: wewμqvi nvi, dc dt = k [A] [B]  k = 1.6  10–4 (0.2  0.2) dm3 mol–1 s –1 = 4.0  10–3 dm3 mol–1 s –1 37. wb‡¤œi wewμqvwUi [NH3] Gi cÖviw¤¢K NbgvÎv 0.75M Ges wewμqvwUi Aa©vqyKvj 30 wgwbU n‡j wewμqvwUi nvi aaæeK KZ n‡e? 2NH3Drcv`b (Product) [KUET 14-15] 7.40  10–4 mol–1min–1 7.40  10–4 mol–1 sec–1 0.023 mol–1min–1 3.85  10–4 mol–1 sec–1 0.0444 mol–1 sec–1 DËi: 7.40  10–4 mol–1 sec–1 e ̈vL ̈v: GKK †`‡L †evSv hvq GwU 2q μg wewμqv| t1 2 = 1 k  a  30 = 1 k  0.75  k = 0.044 mol–1min–1 = 7.4  10–4mol–1 sec–1 38. N2O5 Gi we‡qvRb wewμqvi mwμqY kw3 103.00 kJ mol–1 Ges 25C ZvcgvÎvq †eM aaæeK 2  10–3 sec–1 n‡j 0C ZvcgvÎv‡Z wewμqvi †eM aaæeK KZ n‡e? [KUET 10-11] 4.45  10–5 sec–1 4.45  10–6 sec–1 4.35  10–5 sec–1 4.42  10–7 sec–1 2.67  10–5 sec–1 DËi: 4.45  10–5 sec–1
ivmvqwbK cwieZ©b  Varsity Practice Sheet 5 e ̈vL ̈v: ln k 2  10–3 = Ea R     1 298 – 1 273  ln     k 2  10–3 = 103  1000 8.314     1 298 – 1 273  k = 4.45  10–5 sec–1 39. A  x wewμqvwUi wewμqv nvi n‡jvÑ [RUET 11-12] dA dt & dx dt dx dt & – dA dt dA dt & – dx dt dt dx & – dA dt None of above DËi: dx dt & – dA dt e ̈vL ̈v: †h‡Kvb wewμqv wewμq‡Ki NbvgvÎv n«vm I Drcv‡`i NbgvÎv e„w× cvq| ZvB, – dA dt = dx dt 40. Lime Stone reacts with HCl with the liberation of CO2 gas. Which of the following factor does not have any effect on rate of this reaction? [IUT 14-15] Temperature Pressure Surface Area Concentration DËi: Surface Area 41. The rate of the reaction A + B  Product is k [A] [B]. What is the unit of the rate constant? [IUT 10-11] mol2 dm–6 s –1 mol–2 dm–6 s –1 mol–1 dm3 s –1 mol2 dm–6 s 1 DËi: mol–1 dm3 s –1 e ̈vL ̈v: – dc dt = k [A] [B]  k = mol–1 L s –1 = mol–1 dm3 s –1 wewμqvi μg 42. cÖ_g μ‡gi wewμqvi nvi aaæe‡Ki gvb wb‡Pi †KvbwUi Dci wbf©ikxj bq? [DU 18-19; JU 09-10] ZvcgvÎv wewμqvi Aa©vqy cÖfveK wewμq‡Ki NbgvÎv DËi: wewμq‡Ki NbgvÎv 43. †Kvb wewμqvi NbgvÎv-mgq †jLwPÎ Ab ̈ ̧‡jv †_‡K Avjv`v? [DU 16-17] First order Zero order Second order Fractional order DËi: Zero order e ̈vL ̈v: NbgvÎv mgq NbgvÎv mgq NbgvÎv mgq k~b ̈ μg 1g μg 2q μg 44. cÖ_g μg wewμqv, A  Drcv` Gi †ÿ‡Î wb‡¤œi †jLwPÎmg~n †`Lv‡bv n‡jv| †Kvb †jLwPÎwU Aï×? ([A] = wewμq‡Ki NbgvÎv; t = wewμqvi mgq) [DU 11-12] [A] t In[A] t –d[A] dt [A] In [A]0 [A] t DËi: In[A] t e ̈vL ̈v: 1g μg wewμqvi nvi aaæe‡Ki mgxKiY, k = 1 t ln [A]0 [A]  kt = ln [A]0 – ln [A]  ln [A] = – kt + ln[A]0 hv y = mx + c †iLvi Abyiƒc| t ln [A] ZvB Gi MÖvd n‡e mij‡iLvi mgxKiY, Z‡e g~jwe›`yMvgx bq| 45. A  P cÖ_g μ‡gi wewμqvi †ÿ‡Î wb‡¤œi †Kvb †jLvqb Øviv GKwU g~jwe›`yMvgx mij‡iLv cvIqv hv‡e? [DU 05-06] [A] vs t – d[A] dt vs t log [A] vs t – d[A] dt vs [A] DËi: – d[A] dt vs [A] e ̈vL ̈v: cÖ_g μg wewμqvi †ÿ‡Î, – d[A] dt = k[A] hv y = mx Gi Abyiƒc| ZvB GwU g~jwe›`yMvgx mij‡iLv n‡e|  [A] – d[A] dt 46. GKwU 1g μg wewμqvi 25% mgvß n‡Z 30 min mgq jv‡M| wewμqvwUi Aa©vqyKvj njÑ [BUET 13-14] 60 min 95 min 120 min 72 min DËi: 72 min e ̈vL ̈v: k = 1 t ln C0 C = 1 30 ln 100 75  k = 9.6 × 10–3 min–1  t1 2 = ln 2 k = 72.2 min 47. A  B cÖ_g μg wewμqvi †Kvb †jLwPÎwU mij‣iwLK n‡e? [BUET 13-14, 11-12, 05-06] [A] vs Time ln [A] vs Time 1/[A]2 vs Time 1/[A] vs Time DËi: ln [A] vs Time