반응형 일반화학/[05장] 열화학413 구리의 몰열용량. 226 kJ heat Cu 47.0 kg 12.5℃ 구리의 몰열용량. 226 kJ heat Cu 47.0 kg 12.5℃ If 226 kJ of heat increases the temperature of 47.0 kg of copper by 12.5℃, what is the molar heat capacity of copper? --------------------------------------------------- ▶ 참고: 열용량과 비열 [ https://ywpop.tistory.com/2897 ] --------------------------------------------------- q = C m Δt 로부터, C = q / [m Δt] = (226 kJ) / [47.0 kg × 12.5 ℃] = 0.385 kJ/kg•℃ = 0.385 J.. 2023. 6. 16. 생성엔탈피로 반응엔탈피 계산. CH4(g) 불완전 연소열 생성엔탈피로 반응엔탈피 계산. CH4(g) 불완전 연소열 ΔH 계산. 2CH4(g) + 3O2(g) → 2CO(g) + 4H2O(l) --------------------------------------------------- CH4 불완전 연소 반응식 2CH4(g) + 3O2(g) → 2CO(g) + 4H2O(l) ( 참고 https://ywpop.tistory.com/11595 ) ΔH°_rxn = [생성물들의 ΔH°_f 합] – [반응물들의 ΔH°_f 합] ( 참고 https://ywpop.tistory.com/3431 ) = [2(–110.525) + 4(–285.8)] – [2(–74.9) + 3(0)] ( 참고: ΔH°_f 값 https://en.wikipedia.org/wiki/Standar.. 2023. 6. 16. ΔE 계산. 1.90 atm 8.30 L 2.80 L 350.0 J heat internal energy ΔE 계산. 1.90 atm 8.30 L 2.80 L 350.0 J heat internal energy A piston is compressed from a volume of 8.30 L to 2.80 L against a constant pressure of 1.90 atm. In the process, there is a heat gain by the system of 350.0 J. Calculate the internal energy change. --------------------------------------------------- w = –PΔV ( 참고 https://ywpop.tistory.com/5281 ) = –(1.90) (2.80 – 8.30) = +10.45 atm•L (+.. 2023. 6. 14. gold 55.5 g temp 20℃ 45℃ increase heat gold 55.5 g temp 20℃ 45℃ increase heat What quantity of heat is required to raise the temperature of 55.5 g of gold from 20℃ to 45℃? The specific heat of gold is 0.13 J/g•℃. --------------------------------------------------- ▶ 참고: 열용량과 비열 [ https://ywpop.tistory.com/2897 ] --------------------------------------------------- q = C m Δt = (0.13 J/g•℃) (55.5 g) (45 – 20 ℃) = (0.13) (55.5) (45 – 20.. 2023. 6. 11. Hg 25 g 100. J heat energy temp increase Hg 25 g 100. J heat energy temp increase If 100 J of heat energy is applied to a 25 g sample of mercury, by how many degrees will the temperature of the sample of mercury increase? The specific heat of mercury is 0.14 J/g•℃. --------------------------------------------------- ▶ 참고: 열용량과 비열 [ https://ywpop.tistory.com/2897 ] --------------------------------------------------- q = C m Δt 이므로, Δt =.. 2023. 6. 11. Al 42.7 g 15.2℃ heat energy Al 42.7 g 15.2℃ heat energy What quantity of heat energy must have been applied to a block of aluminum weighing 42.7 g, if the temperature of the block of aluminum increased by 15.2℃? The specific heat of aluminum is 0.89 J/g•℃. --------------------------------------------------- ▶ 참고: 열용량과 비열 [ https://ywpop.tistory.com/2897 ] --------------------------------------------------- q = C m Δt = (0... 2023. 6. 11. –30.℃ 얼음 1.00 mol을 140.℃ 수증기로 (kJ) –30.℃ 얼음 1.00 mol을 140.℃ 수증기로 (kJ) --------------------------------------------------- ▶ 참고: 열용량과 비열 [ https://ywpop.tistory.com/2897 ] --------------------------------------------------- H2O의 몰질량 = 18.015 g/mol 이므로, ▶ 얼음의 비열 = (2.03 J/℃•g) (18.015 g/mol) = 36.6 J/℃•mol ▶ 물의 비열 = (4.18 J/℃•g) (18.015 g/mol) = 75.3 J/℃•mol ▶ 수증기의 비열 = (2.02 J/℃•g) (18.015 g/mol) = 36.4 J/℃•mol ▶ 용융열 = 6.02 kJ/mol.. 2023. 6. 11. 생성엔탈피로 반응엔탈피 계산. CuO 연소열 생성엔탈피로 반응엔탈피 계산. CuO 연소열 CuO(s) + H2(g) → Cu(s) + H2O(l) --------------------------------------------------- ΔH°_rxn = [생성물들의 ΔH°_f 합] – [반응물들의 ΔH°_f 합] ( 참고 https://ywpop.tistory.com/3431 ) = [(0) + (–285.8)] – [(–155.2) + (0)] = –130.6 kJ ---> CuO(s) 1 mol당 반응열이므로, = –130.6 kJ/mol 답: ΔH° = –130.6 kJ/mol 2023. 6. 11. 중화열 계산. 0.500 M HCl + 0.500 M NaOH 1.00×10^2 mL 중화열 계산. 0.500 M HCl + 0.500 M NaOH 1.00×10^2 mL 일정 압력 열량계에서 0.500 M HCl 1.00×10^2 mL와 0.500 M NaOH 1.00×10^2 mL를 혼합하였다. HCl 용액과 NaOH 용액의 초기 온도는 22.50℃이었고, 혼합 용액의 최종 온도는 25.86℃이었다. 몰당 중화열을 계산하라. 용액의 밀도와 비열은 물의 값과 같다고 가정한다. (각각 1.00 g/mL, 4.184 J/g•℃) 열량계의 열용량은 무시한다. --------------------------------------------------- 1.00×10^2 mL = 100. mL = 100. g 1) 용액이 흡수한 열 q = C m Δt ( 참고 https://ywpop.tisto.. 2023. 6. 10. ΔH° = –484 kJ 0.25 mol O2 0.50 mol H2 –5.6 L PV work ΔE ΔH° = –484 kJ 0.25 mol O2 0.50 mol H2 –5.6 L PV work ΔE The reaction between hydrogen and oxygen to yield water vapour has ΔH° = –484 kJ. 2H2(g) + O2(g) → 2H2O(g) ... ΔH° = –484 kJ How much PV work is done and what is the value of ΔE (in kJ) for the reaction of 0.50 mol of H2 and 0.25 mol of O2 at atmospheric pressure if the volume change is –5.6 L? ----------------------------------------------.. 2023. 6. 9. 이전 1 2 3 4 5 6 7 8 ··· 42 다음 반응형