반응형 일반화학/[05장] 열화학384 0.0℃ 1.00 atm 39.1 mol He volume 876 L 38.0℃ 998 L q w ΔE 20.8 J/℃ 0.0℃ 1.00 atm 39.1 mol He volume 876 L 38.0℃ 998 L q w ΔE 20.8 J/℃ A balloon filled with 39.1 moles of helium has a volume of 876 L at 0.0℃ and 1.00 atm pressure. At constant pressure, the temperature of the balloon is increased to 38.0℃, causing the balloon to expand to a volume of 998 L. Calculate q, w, and ΔE for the helium in the balloon. The molar heat capacity for helium gas is 20.8 J/℃•mo.. 2021. 7. 8. 52.5 J heat 0.500 atm 58.0 L 102.5 J volume 52.5 J heat 0.500 atm 58.0 L 102.5 J volume As a system increases in volume, it absorbs 52.5 J of energy in the form of heat from the surroundings. The piston is working against a pressure of 0.500 atm. The final volume of the system is 58.0 L. What was the initial volume of the system if the internal energy of the system decreased by 102.5 J? --------------------------------------------------- .. 2021. 7. 7. 정반응 활성화에너지는 125 kJ/mol, ΔE는 –216 kJ/mol 정반응 활성화에너지는 125 kJ/mol, ΔE는 –216 kJ/mol A + B → C + D 반응에 대한 활성화 에너지는 125 kJ/mol이고, 반응의 ΔE는 –216 kJ/mol이다. 역반응의 활성화 에너지는 얼마인가? --------------------------------------------------- ΔE < 0 이므로, 발열 반응. ( 참고 https://ywpop.tistory.com/5242 ) [그림] E_a1 = 정반응의 활성화에너지. E_a2 = 역반응의 활성화에너지. 역반응의 활성화 에너지 = 216 + 125 = 341 kJ/mol [키워드] 역반응의 활성화에너지 기준문서 2021. 6. 21. 규소 45.0 g을 6.0℃ 올리기 위해서는 192 J이 필요 규소 45.0 g을 6.0℃ 올리기 위해서는 192 J이 필요 규소 45.0 g을 6.0℃ 올리기 위해서는 192 J이 필요하다. 규소의 몰열용량[J/(mol•℃)]은 얼마인가? --------------------------------------------------- ▶ 참고: 열용량과 비열 [ https://ywpop.tistory.com/2897 ] --------------------------------------------------- Si의 몰질량 = 28.09 g/mol 45.0 g / (28.09 g/mol) = 1.602 mol ( 참고 https://ywpop.tistory.com/7738 ) 192 J / [(1.602 mol) (6.0℃)] = 19.98 J/(mol•℃) 2021. 6. 16. 1000 g water 0.7521 g benzoic acid 3.60℃ –26.42 kJ/g 1000 g water 0.7521 g benzoic acid 3.60℃ –26.42 kJ/g When 0.7521 g of benzoic acid was burned in a calorimeter containing 1000. g of water, a temperature rise of 3.60℃ was observed. What is the heat capacity of the bomb calorimeter, excluding the water? The heat of combustion of benzoic acid is –26.42 kJ/g. --------------------------------------------------- ▶ 참고: 열용량과 비열 [ https://ywpop.tistory.. 2021. 6. 14. 117.82℃ 30.14 g metal 18.44℃ water 120.0 mL 0.474 J/g•℃ 117.82℃ 30.14 g metal 18.44℃ water 120.0 mL 0.474 J/g•℃ A 30.14 g stainless steel ball bearing at 117.82℃ is place in a constant pressure calorimeter containing 120.0 mL of water at 18.44℃. If the specific heat of the ball bearing is 0.474 J/g•℃, calculate the final temperature of the water. Assume the calorimeter to have negligible heat capacity. ------------------------------------------------.. 2021. 6. 9. 1.07 M HCHO2 75.0 mL 22.4℃ 3.20 M NaOH 25.0 mL 23.4℃ 1.07 M HCHO2 75.0 mL 22.4℃ 3.20 M NaOH 25.0 mL 23.4℃ 반응열을 측정하기 위해 1.07 M HCHO2 75.0 mL를 22.4℃의 커피컵 열량계에 넣고 22.4℃의 3.20 M NaOH 25.0 mL를 가했다. 혼합물을 온도계로 빠르게 교반했더니, 온도가 23.4℃까지 상승하였다. In the reaction between formic acid (HCHO2) and sodium hydroxide, water and sodium formate (NaCHO2) are formed. To determine the heat of reaction, 75.0 mL of 1.07 M HCHO2 was placed in a coffee cup calorimeter at a tem.. 2021. 6. 9. ΔH = +25.7 kJ 50.0 g NH4NO3 water 125 mL 4.18 J 25.0℃ ΔH = +25.7 kJ 50.0 g NH4NO3 water 125 mL 4.18 J 25.0℃ Instant cold packs are effective because the dissolution of solid NH4NO3 is endothermic: NH4NO3(s) + H2O(l) → NH4NO3(aq) ... ΔH = +25.7 kJ/mol What is the final temperature in a squeezed cold pack that contains 50.0 g of NH4NO3 dissolved in 125 mL of water? Assume a specific heat of 4.18 J/g•℃ for the solution, an initial temperature of 25.0℃.. 2021. 6. 8. 0℃ 얼음 5 kg을 0℃ 물로 만드는데 필요한 에너지 0℃ 얼음 5 kg을 0℃ 물로 만드는데 필요한 에너지 얼음의 융해열: 3.33×10^5 J/kg --------------------------------------------------- ▶ 참고: 열용량과 비열 [ https://ywpop.tistory.com/2897 ] q = C m = (3.33×10^5 J/kg) (5 kg) = 1665000 J = 1665 kJ [ 관련 예제 https://ywpop.tistory.com/18249 ] –20℃ 얼음 1 kg을 100℃ 수증기로 (kJ) 2021. 6. 7. ΔH 계산. H2 + Cl2 → 2HCl 결합엔탈피로 ΔH 계산. H2 + Cl2 → 2HCl 결합엔탈피로 결합엔탈피로 반응엔탈피 계산 다음 반응의 엔탈피 변화를 계산하시오. H2(g) + Cl2(g) → 2HCl(g) 단, 이원자 분자들의 결합엔탈피는 다음과 같다. H2: 436.4 kJ/mol Cl2: 242.7 kJ/mol HCl: 431.9 kJ/mol --------------------------------------------------- ▶ 참고: 반응 엔탈피를 계산하는 2가지 방법 [ https://ywpop.tistory.com/13107 ] ΔH_rxn = Σ(절단 결합의 결합 엔탈피) – Σ(생성 결합의 결합 엔탈피) ( 참고 https://ywpop.tistory.com/3729 ) = [(436.4) + (242.7)] – [2(4.. 2021. 6. 7. 이전 1 ··· 20 21 22 23 24 25 26 ··· 39 다음 반응형
