Tiêu đề 4. Che. Engg. 1st Paper-Practice Sheet-24 (With Solve).pdf ✅

ivmvqwbK cwieZ©b  Engineering Practice Sheet ............................................................................................................... 1 WRITTEN weMZ mv‡j BUET-G Avmv cÖkœvejx 1. (i) 2C4H10 + 7O2  2C4H2O3 + 8H2O (ii) 2SO2 + O2  2SO3 (iii) Cr2O3 + 3CO  2Cr + 3CO2 Dch©y3 wewμqv ̧‡jv wMÖb †Kwgw÷ai μg Abymv‡i mvRvI: [BUET 22-23] [Easy] mgvavb: (i) GUg BKbwg (%A/E) = Kvw•ÿZ Drcv‡`i †gvU †gvj  ms‡KZ fi  100% mKj Drcv` ev wewμq‡Ki †gvj msL ̈v mn ms‡KZ f‡ii mgwó = 2 × C4H2O3 2 × C4H10 + 7 × O2 × 100% = 2 × 98 (2 × 58) + (7 × 32) × 100% = 57.64% (ii) bs wewμqvi †ÿ‡Î, %AE = 2 × SO3 2 × SO2 + O2 × 100% = 2 × 80 2 × 64 + 32 × 100% = 100% (iii) bs wewμqvi †ÿ‡Î, %AE = 2 × Cr Cr2O3 + 3 × CO × 100% = 2 × 52 × 100% (2 × 52 + 48) + (3 × 28) = 44.067% wMÖb †Kwgw÷ai μg ii > i > iii 2. 2H2S + Heat  2H2(g) + S2(g) mvg ̈ve ̄ vq †Kvb w`‡K cwieZ©b n‡e (Wv‡b/ev‡g)? (i) ZvcgvÎv evov‡j (ii) Pvc evov‡j (iii) S †hvM Ki‡j (iv) H2S †hvM Ki‡j (v) S AcmviY Ki‡j [BUET 22-23] [Easy] mgvavb: mvg ̈ve ̄ vi cwieZ©b (i) ZvcgvÎv evov‡j Wv‡b (ii) Pvc evov‡j ev‡g (iii) S †hvM Ki‡j ev‡g (iv) H2S †hvM Ki‡j Wv‡b (v) S AcmviY Ki‡j Wv‡b 3. 1 wjUvi cvwb‡Z 53.45 mg NH4Cl `aexf~Z Kiv n‡q‡Q Ges `ae‡Yi pH Gi gvb 8 Kiv n‡q‡Q| GB `ae‡Y [NH+ 4 ] I [NH3] Gi NbZ¡ (M) wbY©q Ki| [NH+ 4 Gi Rb ̈ Ka = 10–9.26] [BUET 21-22] [Hard] mgvavb: NH+ 4 Gi Ka = 10–9.26  pKa = 9.26 NH3 Gi pKb = 14 – 9.26 = 4.74 pH = 8  pOH = 14 – 8 = 6 pOH = pKb + log nNH4Cl nNH3  6 = 4.74 + log nNH4Cl nNH3  nNH4Cl nNH3 = 101.26 = 18.2  53.45 × 10–3 14 + 4 + 35.5 nNH3 = 18.2 nNH3 = 5.5 × 10–5 mol  [NH3] = 5.5 × 10–5 M [†h‡nZz cvwb 1 L] Ges  [NH+ 4 ] = 53.45 × 10–3 14 + 4 + 35.5 1 = 9.99 × 10–4 M 4. †Kv‡bv GKwU wewμqvi ZvcgvÎv 427C †_‡K 527C G DbœxZ Kivi d‡j wewμqvi nvi aaæeK wØ ̧Y n‡q †Mj| wewμqvwUi mwμqY kw3 (kJ/mol GK‡K) KZ? [BUET 20-21] [Medium] mgvavb: †`Iqv Av‡Q, T1 = 427C = 700 K T2 = 527C = 800 K †h‡nZz nvi aaæeK wØ ̧Y nq, k2 = 2k1 ln     k2 k1 = Ea R     1 T1 – 1 T2  ln     2k1 k1 = Ea R     1 700 – 1 800  Ea = ln2  R (700–1 – 800–1 )  Ea = 32.22 kJ mol–1 5. wb‡¤œv3 ivmvqwbK wewμqvmg~‡ni mvg ̈ve ̄ vi Pvc n«vm K‡i M ̈vmxq AvqZb e„w× Ki‡j wewμqvi wgkÖ‡Y Drcv‡`i †gvj msL ̈vi Kx cwieZ©b n‡e e ̈vL ̈v Ki| [BUET 20-21] [Easy] (i) CaO(s) + CO2(g) ⇌ CaCO3(s) (ii) 3Fe(s) + 4H2O(g) ⇌ Fe3O4(s) + 4H2(g) mgvavb: (i) M ̈vmxq wewμqK I Drcv‡`i †gvj msL ̈vi cv_©K ̈ n = Drcv` – wewμqK = 0 – 1 = – 1 AZGe, Drcv‡`i †gvj msL ̈v n«vm cv‡e| (ii) n = Drcv` – wewμqK = 4 – 4 = 0 †gvj msL ̈v AcwiewZ©Z _vK‡e, ZvB Pv‡ci †Kv‡bv cÖfve †bB|
2 ....................................................................................................................................  Chemistry 1st Paper Chapter-4 6. GKwU dv‡b©‡m 35% CO2, 65% CO M ̈vm wgkÖY e ̈envi K‡i 700C G w÷j‡K Zvc w`‡j Gi Dci gwiPv co‡e wKbv Zv wbY©q Ki| †`Iqv Av‡Q: Fe(s) + CO2(g) = FeO(s) + CO(g); K = 1.43| [BUET 20-21] [Medium] mgvavb: Fe(s) + CO2(g) = FeO(s) + CO(g) QP = PCO PCO2 = 0.65 0.35 = 1.86 Avgiv Rvwb, K > Q  m¤§yLgyLx K < Q  cðvrgyLx K = Q  mvg ̈ve ̄’v QP = 1.86 > K A_©vr, wewμqvwU cðvrgyLx| myZivs, FeO •Zwi n‡e bv, gwiPvI co‡e bv| 7. 1 L AvqZ‡bi GKwU cv‡Î hLb 0.1 mol PCl5 †K DËß Kiv nq, mvg ̈wgkÖ‡Yi †gvU Pvc nq 4.38  105 Nm–2 | ZvcgvÎv T = 450 K G mvg ̈aaæeK, (KP) Gi gvb wbY©q Ki| [BUET 19-20] [Medium] mgvavb: PCl5 ⇌ PCl3 + Cl2 ïiæ‡Z: 0.1 0 0 mvg ̈ve ̄’vq: (0.1 – )    n = 0.1 –  +  +  = (0.1 + ) Avgiv Rvwb, PV = nRT  n = PV RT n = 4.38 105  1  10–3 8.314  450 = 0.11707 mol kZ©g‡Z, 0.1 +  = 0.11707 mol  = 0.01707  KP = PPCl3 .PCl2 PPCl5 =      0.1 +       0.1 +  P  total     0.1 –  0.1 +  Ptotal =      2 0.1 +  Ptotal (0.1 – ) = 13145.70 Nm–2 = 0.129738 atm 8. wb‡Pi d¬zBW ̧‡jvi pH gv‡bi mxgv wjL| [BUET 19-20] [Easy] i. gy‡Li jvjv ii. cvK ̄ wji im mgvavb: i. 6.2 – 7.4 ii. 1.5 – 3.5 9. UvBUvwbqvg `ywU wfbœ wfbœ c×wZ Øviv AvKwiK †_‡K wb®..vkb Kiv hvq- (i) AwaKZi mwμq avZzi e ̈envi, TiO2 + 2Mg  Ti + 2MgO (ii) AvKwi‡Ki Zwor we‡kølY, TiO2  Ti + O2; Kvw•ÿZ Drcv‡` wewμqK cigvYyi me©vwaK Dcw ̄ wZi aviYv e ̈envi K‡i Dc‡ii †Kvb c×wZwU wMÖbvi Zv wbY©q Ki? [Ti = 47.88 and Mg = 24.3] [BUET 18-19] [Easy] mgvavb: kZKiv GUg BKbwg, %AE = Kvw•ÿZ Drcv‡`i †gvU †gvj msL ̈v × ms‡KZ fi × 100% me wewμq‡Ki ev Drcv‡`i †gvU †gvj msL ̈v mn ms‡KZ f‡ii mgwó (i) bs wewμqvi †ÿ‡Î, % AE = 47.88 47.88 + 80.6  100% = 37.27% (ii) bs wewμqvi †ÿ‡Î, % AE = 47.88 47.88 + 32  100% = 59.94% myZivs (ii) bs c×wZwU wMÖbvi| 10. GK †gvj A ̈vwmwUK GwmW I GK †gvj †mvwWqvg A ̈vwm‡UU m¤^wjZ GK wjUvi `ae‡Y 4 g NaOH `aexf~Z Kiv n‡jv| Drcvw`Z wgkÖYwUi pH wbY©q Ki| [A ̈vwmwUK Gwm‡Wi we‡qvRb aaæeK = 1.8  10–5 ] [BUET 19-20] [Medium] mgvavb: pH = pKa + log nCH3COONa nCH3COOH = – log(1.8  10–5 ) + log 1 + 4 40 1 – 4 40 = 4.832 11. Møy‡KvR Ges d«z‡±vR Rjxq `ae‡Y wb¤œiƒ‡c mvg ̈ve ̄ vq _v‡K| d«z‡±vR ⇌ Møy‡KvR GKRb QvÎ 0.244 M d«z‡±vR `aeY •Zwi Kij| 25C ZvcgvÎvq mvg ̈ve ̄ vq Gi NbZ¡ K‡g 0.113 M G cwiYZ nj| (a) wewμqvwUi mvg ̈aaæeK wbY©q Ki| [BUET 18-19] [Easy] (b) kZKiv KZ fvM d«z‡±vR Møy‡Kv‡R cwiYZ nj| mgvavb: (a) C6H12O6 (d«z‡±vR) ⇌ C6H12O6 (Møy‡KvR) ïiæ‡Z: 0.244 M 0 M mvg ̈ve ̄’vq: 0.113 M (0.244 – 0.113) = 0.131 M  mvg ̈aaæeK, KC = [Møy‡KvR] [d«z‡±vR] = 0.131 0.113 = 1.16 (b) mvg ̈ve ̄’vq d«z‡±vR Møy‡Kv‡R cwiYZ nIqvi cwigvY,  = 0.131 0.244  12. A ̈v‡gvwbqv Aw·‡R‡bi mv‡_ wb‡¤œi mgxKiY Abyhvqx wewμqv K‡i: 4NH3(g) + 5O2(g) = 4NO(g) + 6H2O(g) †Kv‡bv gyn~‡Z© A ̈v‡gvwbqv 0.24 molL–1 s –1 nv‡i wewμqv Ki‡j wewμqvwUi nvi mgxKiYwU wjL Ges H2O(g) Drcv`‡bi nvi wbY©q Ki| [BUET 17-18] [Hard] mgvavb: 4NH3 + 5O2 = 4NO + 6H2O wewμqvi nvi = – 1 4 d[NH3] dt = – 1 5 d[[O2] dt = 1 4 d[NO] dt = 1 6 d[H2O] dt d[NH3] dt = – 0.24 mol L–1 s –1  d dt [H2O] = – 6 4 d[NH3] dt = – 6 4  (– 0.24) mol L–1 s –1 = 0.36 mol L–1 s –1
ivmvqwbK cwieZ©b  Engineering Practice Sheet .............................................................................................................. 3 13. 400 mL 0.1 M NaOH Gi mv‡_ 600 mL 0.2 M A ̈vwmwUK GwmW †hvM K‡i GKwU evdvi `aeY cÖ ̄‧Z Kiv nj| evdvi `ae‡Yi pH KZ? [A ̈vwmwUK Gwm‡Wi Rb ̈ pKa = 4.76][BUET 17-18] [Medium] mgvavb: 400 mL 0.1 M NaOH = 0.04 mol NaOH 600 mL 0.2 M CH3COOH = 0.12 mol CH3COOH 0.04 mol NaOH, 0.04 mol CH3COOH Gi mv‡_ wewμqv K‡i 0.04 mol CH3COONa •Zwi K‡i| Aewkó CH3COOH Gi cwigvY (0.12 – 0.04) = 0.08 mol pH = pKa + log [Salt] [Acid] = 4.76 + log [0.04] [0.08] = 4.46 14. GKwU mvi KviLvbv †_‡K A ̈v‡gvwbqv M ̈vm wbtmwiZ n‡q cv‡ki grm ̈ Pv‡li GKwU cyKz‡ii ̄^”Q cvwb‡Z `aexf~Z nj| hw` `aexf~Z A ̈v‡gvwbqvi NbgvÎv 0.01 M Ges we‡qvRb aaæeK Kb = 1.8  10–5 nq Zvn‡j cyKz‡ii cvwbi pH KZ n‡e? cyKziwU grm ̈ Pv‡li Dchy3 n‡e wK? [BUET 16-17] [Medium] mgvavb: Kb = 1.8  10–5 ; C = 0.01 M Avgiv Rvwb, pOH = – log(c)  pOH = – log   Kb  C × C      = Kb C  pOH = – log( KbC) = 3.372 pH = 14 – 3.372 = 10.628 gv‡Qi Rxeb avi‡Yi Rb ̈ cvwbi m‡ev©Ëg pH n‡”Q (6.7 – 8.6)| ZvB cyKziwU grm ̈ Pv‡li Dchy3 n‡e bv| 15. wb‡Pi QKwU mwVK pH range w`‡q c~iY Ki: [BUET 16-17] [Easy] Subject pH Range Human Blood (gvbe‡`‡ni i3) Pottery soil (g„r wk‡íi gvwU) Leather tanning (Pvgov U ̈vwbs) Toilet soap (†Mvm‡ji mvevb) mgvavb: Subject pH Range Human Blood (gvbe †`‡ni i3) 7.35 – 7.45 Pottery soil (g„r wk‡íi gvwU) 6 – 6.5 Leather tanning (Pvgov U ̈vwbs) 4.0 – 4.5 Toilet soap (†Mvm‡ji mvevb) 7 – 8 16. nvB‡Wav‡Rb Avq‡bi 5.6  10–2 M NbgvÎv wewkó 500 mL `ae‡Y Kgjv‡jeyi im wgwkÖZ K‡i d‡ji i‡mi GKwU 1 L wgkÖY •Zwi Kiv n‡jv| d‡ji i‡mi wgkÖ‡Yi pH KZ n‡e? d‡ji i‡mi wgkÖYwU cvb‡hvM ̈ n‡e wK? [Kgjv‡jeyi i‡mi NbgvÎv 4.4 × 10–2 M] [BUET 16-17; CUET 10-11] [Medium] mgvavb: NbgvÎv = †gvj msL ̈v wjUv‡i AvqZb  5.6  10–2 = n1 500 1000  n1 = 5.6  10–2 × 500 1000 = 0.028 mol Ges n2 = 4.4  10–2  500 1000 = 0.022 mol †gvU †gvj = (n1 + n2) = 0.05 mol †gvU AvqZb = (500 + 500) mL = 1 L  [H+ ] = 0.05 1 = 0.05 M  pH = – log[H+ ] = – log(0.05) = 1.3; hv AwaK A¤øxq|  d‡ji i‡mi wgkÖYwU cvb‡hvM ̈ n‡e bv| 17. M ̈vmxq Ae ̄ vq mvB‡K¬v‡cÖv‡cb •Zix GKwU cÖ_g μg wewμqv, hvi nvi aaæeK 500C G 6.7  10-4 s –1 | (K) mvB‡K¬v‡cÖv‡cb Gi cÖv_wgK NbgvÎv hw` 0.25 M nq Z‡e 8.8 min ci Gi NbgvÎv KZ n‡e? (L) mvB‡K¬v‡cÖv‡cb Gi NbgvÎv 0.25 M †_‡K n«vm †c‡q 0.15 M n‡Z KZ mgq jvM‡e? [BUET 16-17] [Hard] mgvavb: 500C ZvcgvÎvq, k = 6.7  10–4 s –1 (K) k = 2.303 t log a a – x log(a) – log(a – x) = kt 2.303  log (a – x) = loga – kt 2.303  log (a – x) = log(0.25) – 6.7  10–4  8.8  60 2.303  log (a – x) = – 0.7557  a – x = 0.1755 M (L) k = 2.303 t log a a – x 6.7 × 10–4 = 2.303 t log 0.25 0.15  t = 762.565 sec = 12.709 min 18. bvB‡Uav‡Rb †c›UvA·vB‡Wi we‡qvRb (N2O5  N2O4 + O2) Gi nvi aaæe‡Ki gvb 25C I 65C ZvcgvÎvq h_vμ‡g 3.46  10–5 I 4.48  10–3 | G wewμqvi mwμqY kw3i gvb KZ n‡e? [BUET 15-16; 08-09] [Medium] mgvavb: †`Iqv Av‡Q, T1 = 298 K; T2 = 338 K k1 = 3.46 × 10–5 ; k2 = 4.48 × 10–3 Avgiv Rvwb, log k2 k1 = Ea 2.303R     T2 – T1 T1T2  log     4.48  10–3 3.46  10–5 = Ea 2.303  8.314      338 – 298 298  338  2.1122 = Ea  (2.0734  10–5 )  Ea = 101871.3225 J = 101.87 kJ mol–1
4 ....................................................................................................................................  Chemistry 1st Paper Chapter-4 19. A ̈vwmwUK GwmW evdv‡ii pH = 4.6 †c‡Z n‡j A¤ø I je‡Yi AbycvZ †ei K‡i| [CH3COOH Gi Ka = 1.8  10–5 ] [BUET 15-16; KUET 14-15] [Easy] mgvavb: pH = 4.6; Ka = 1.8  10–5  pKa = – log Ka = – log (1.8 × 10–5 ) = 4.745 †nÛvimb n ̈v‡mjevL mgxKiY g‡Z, pH = pKa + log [salt] [acid]  log [salt] [acid] = pH – pKa  log [salt] [acid] = 4.6 – 4.745  log [salt] [acid] = – 0.145  [salt] [acid] = 0.72  [acid] [salt] = 1.39 20. H2 + Br2 = 2HBr wewμqvwU GKwU 0.250 L cv‡Î m¤úbœ Kiv nj| 0.01 s G Br2 Gi cwigv‡Yi cwieZ©b – 0.001 mol n‡j wewμqvwUi nvi wbY©q Ki| [BUET 14-15 ; KUET 03-04] [Medium] mgvavb: wewμqvi nvi dC dt = – – 0.001 0.250  0.01 = 0.4 mol L–1 s –1 1. N2O5 Gi we‡qvRb wewμqvi mwμqY kw3 103.05 kJ mol–1 | 0C Ges 25C ZvcgvÎvq wewμqvwUi †eM aaæe‡Ki Abycv‡Zi gvb wbY©q Ki| [BUET 13-14] [Medium] mgvavb: ln     k2 k1 = Ea R     1 T1 – 1 T2  ln     k2 k1 = 103.05  103 8.314     1 273 – 1 298  ln     k2 k1 = 3.8089      k2 k1 = e3.8089      k2 k1 = 45.1008  k2 : k1 = 45.1008 : 1 21. (i) ivmvqwbK mvg ̈ve ̄ vi kZ© ̧wj wjL| (ii) 25C ZvcgvÎvq mvg ̈ve ̄ vq N2O4 Gi we‡qvR‡bi wgkÖ‡Y N2O4 Gi AvswkK Pvc 0.75 atm Ges wewμqvwUi KP = 8.33 × 10–2 atm| wewμqvwUi KC Ges NO2 Gi AvswkK Pvc wbY©q Ki| [BUET 13-14] [Medium] mgvavb: (i) ivmvqwbK mvg ̈ve ̄’vi kZ©: wewμqvi Am¤ú~Y©Zv, cÖfve‡Ki f~wgKvnxbZv, Dfq w`K n‡Z myMg ̈Zv, mv‡g ̈i ̄’vqxZ¡| (ii) N2O4 ⇌ 2NO2  KP = (PNO2 ) 2 PN2O4  PNO2 = Kp × PN2O4 = 8.33 × 10–3 × 0.75 = 0.25 atm n = 2 – 1 = 1 GLb, KP = KC (RT)n  KP = KC (RT)1  KC = KP RT  KC = 8.33 × 10–2 (0.0821 × 298) 1  KC = 3.404 × 10–3 mol/L 22. HNO3 Ges H3PO4 Gi g‡a ̈ †KvbwU AwaK kw3kvjx? †Zvgvi Dˇii h_v_©Zv e ̈vL ̈v Ki| [BUET 13-14] [Easy] mgvavb: HNO3 Ges H3PO4 Gi g‡a ̈ HNO3 AwaK kw3kvjx|   +5 +5 Gwm‡Wi †K›`axq cigvYyi RviY gvb mgvb n‡j GwmW ̧‡jvi g‡a ̈ †hwUi †K›`axq cigvYyi AvKvi †QvU A_©vr ewnt ͇̄i PvR©NbZ¡ †ewk Zvi ZxeaZv †ewk n‡e| G‡ÿ‡Î N Gi AvKvi P †_‡K †QvU nIqvq HNO3 AwaK kw3kvjx| 23. 4.60 pH Gi GKwU evdvi `aeY •Zwi Ki‡Z n‡e| GRb ̈ GKwU weKv‡i 0.01 M A ̈vwmwUK Gwm‡Wi 10.0 mL `aeY †bqv n‡jv| G‡Z 0.01 M †mvwWqvg A ̈vwm‡U‡Ui KZ AvqZb (mL) `aeY †hvM Ki‡Z n‡e? [pKa = 4.75] [BUET 12-13] [Easy] mgvavb: pH = pKa + log [jeY] [A¤ø]  pH – pKa = log [jeY] [A¤ø]  – 0.15 = log [jeY] [A¤ø]  [0.01  V] [0.01  10] = anti log (– 0.15)  V = 7.079 mL 24. WvBbvB‡Uav‡Rb †UUav·vB‡Wi we‡qvRbwU wb¤œiƒc: N2O4(g) ⇌ 2NO2(g) mvg ̈ve ̄ vq 25C ZvcgvÎvq M ̈vm `ywUi AvswkK Pvc h_vμ‡g, PN2O4 = 0.69 atm Ges PNO2 = 0.31 atm. A. mvg ̈aaæeK KP I KC wbY©q Ki| B. GKB ZvcgvÎvq WvBbvB‡Uav‡Rb †UUav·vB‡Wi we‡qvRb gvÎv wbY©q Ki| [BUET 08-09] [Medium] mgvavb: A. N2O4 ⇌ 2NO2  KP = (PNO2) 2 PN2O4 = 0.139 atm KP = KC(RT)n C KP (RT) n  0.139 (0.0821  298) [ n = 2 – 1 = 1] = 5.68  10–3