Tiêu đề Chemical Change Engg Question Bank Solution (HSC 26).pdf ✅

ivmvqwbK cwieZ©b  Engineering Practice Sheet ............................................................................................................... 1 WRITTEN weMZ mv‡j BUET-G Avmv cÖkœvejx 1. PCl3  Cl2 ⇌ PCl5 Gi Kc  49. PCl3 I Cl2 Gi cÖ‡Z ̈KwU 0.5 mole wb‡q 5 L AvqZ‡bi cv‡Î ivL‡j mvg ̈ve ̄ vq cÖ‡Z ̈‡Ki †gvj msL ̈v KZ? [BUET 23-24] mgvavb: PCl3 + Cl2 PCl5 ïiæ‡Z: 0.5 0.5 0 mvg ̈ve ̄’vq: 0.5 –  0.5 –   Kc =  5     0.5 –  5 2  49 = 5 (0.5 – ) 2  49 2 – 54  12.25 = 0   = 0.78, 0.32 [   0.78]   = 0.32 mvg ̈wgkÖ‡Y, nPCl3 = 0.5 – 0.32 = 0.18 mol nCl2 = 0.5 – 0.32 = 0.18 mol nPCl5 = 0.32 mol 2. PCl3(g) + Cl2(g) ⇌ PCl5(g) wewμqvi Kc = 49, 0.5 mol PCl3 I Cl2 wb‡q wewμqv Ki‡j mvg ̈ve ̄ vq †gvj msL ̈v KZ? [V = 5 L] [BUET 23-24] mgvavb: awi, mvg ̈ve ̄’vq x †gvj PCl3 I Cl2 we‡qvwRZ nq| PCl3 + Cl2 ⇌ PCl5 ïiæ‡Z (mol) : 0.5 0.5 0 mvg ̈ve ̄’vq (mol) : (0.5 – x) (0.5 – x) x myZivs, KC = [PCl5] [PCl3]  [Cl2] = x 5     0.5 – x 5 2 [∵ S = n V ]  x = 0.3195 mol [∵ KC = 49] AZGe, mvg ̈ve ̄’vq †gvj msL ̈v, nPCl 3 = nCl 2 = (0.5 – 0.3195) = 0.1805 mol I nPCl 5 = 0.3195 mol (Ans.) 3. (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 4. 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 5. 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
2 ....................................................................................................................................  Chemistry 1st Paper Chapter-4 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 6. †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 7. 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| 8. 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| 9. 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 10. 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 11. 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

4 ....................................................................................................................................  Chemistry 1st Paper Chapter-4  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| 19. 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 20. 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 21. 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 22. 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 23. (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