(1)已知热化学方程式H2O(g)====H2(g) + ½O2(g) △H=+241.8kJ/molH2(g) + ½O2(g)====H2O(l ) △H=-285.8kJ/mol当1g液态水变为水蒸气时,其热量变化是多少?(2)已知Fe2O3(s)+ 3/2 C(s)====3/2 CO2(g) + 2Fe(s) △H1
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![(1)已知热化学方程式H2O(g)====H2(g) + ½O2(g) △H=+241.8kJ/molH2(g) + ½O2(g)====H2O(l ) △H=-285.8kJ/mol当1g液态水变为水蒸气时,其热量变化是多少?(2)已知Fe2O3(s)+ 3/2 C(s)====3/2 CO2(g) + 2Fe(s) △H1](/uploads/image/z/1437187-67-7.jpg?t=%EF%BC%881%EF%BC%89%E5%B7%B2%E7%9F%A5%E7%83%AD%E5%8C%96%E5%AD%A6%E6%96%B9%E7%A8%8B%E5%BC%8FH2O%28g%29%3D%3D%3D%3DH2%28g%29+%2B+%26%23189%3BO2%EF%BC%88g%EF%BC%89+%E2%96%B3H%3D%2B241.8kJ%2FmolH2%28g%29+%2B+%26%23189%3BO2%28g%29%3D%3D%3D%3DH2O%EF%BC%88l+%29+%E2%96%B3H%3D-285.8kJ%2Fmol%E5%BD%931g%E6%B6%B2%E6%80%81%E6%B0%B4%E5%8F%98%E4%B8%BA%E6%B0%B4%E8%92%B8%E6%B0%94%E6%97%B6%2C%E5%85%B6%E7%83%AD%E9%87%8F%E5%8F%98%E5%8C%96%E6%98%AF%E5%A4%9A%E5%B0%91%3F%EF%BC%882%EF%BC%89%E5%B7%B2%E7%9F%A5Fe2O3%28s%29%2B+3%2F2+C%28s%29%3D%3D%3D%3D3%2F2+CO2%28g%29+%2B+2Fe%28s%29+%E2%96%B3H1)
(1)已知热化学方程式H2O(g)====H2(g) + ½O2(g) △H=+241.8kJ/molH2(g) + ½O2(g)====H2O(l ) △H=-285.8kJ/mol当1g液态水变为水蒸气时,其热量变化是多少?(2)已知Fe2O3(s)+ 3/2 C(s)====3/2 CO2(g) + 2Fe(s) △H1
(1)已知热化学方程式
H2O(g)====H2(g) + ½O2(g) △H=+241.8kJ/mol
H2(g) + ½O2(g)====H2O(l ) △H=-285.8kJ/mol
当1g液态水变为水蒸气时,其热量变化是多少?
(2)已知
Fe2O3(s)+ 3/2 C(s)====3/2 CO2(g) + 2Fe(s) △H1=+234.1kJ/mol
C(s) + O2(g)====CO2(g) △H2=-393.5kJ/mol
则2Fe(s) + 3/2 O2(g)====Fe2O3(s)的△H3是多少?
(1)已知热化学方程式H2O(g)====H2(g) + ½O2(g) △H=+241.8kJ/molH2(g) + ½O2(g)====H2O(l ) △H=-285.8kJ/mol当1g液态水变为水蒸气时,其热量变化是多少?(2)已知Fe2O3(s)+ 3/2 C(s)====3/2 CO2(g) + 2Fe(s) △H1
(1) H2O(g)====H2(g) + ½O2(g) △H1=+241.8kJ/mol ①
H2(g) + ½O2(g)====H2O(l ) △H2=-285.8kJ/mol ②
H2O(g)====H2O(l ) △H3 ③
当1g液态水变为水蒸气时,其热量变化是多少?
①+②=③则 H2O(g)====H2O(l ) △H3==+241.8kJ/mol -285.8kJ/mol= -44kJ/mol
所以 H2O(l)====H2O(g ) △H4=44kJ/mol 因此1g水=1/18mol 所以1g水转化成水蒸汽需要吸收44kJ/mol*1/18mol=2.44KJ能量
(2)Fe2O3(s)+ 3/2 C(s)====3/2 CO2(g) + 2Fe(s) △H1=+234.1kJ/mol ①
C(s) + O2(g)====CO2(g) △H2=-393.5kJ/mol ②
Fe(s) + 3/2 O2(g)====Fe2O3(s) △H3=?③
-①+3/2②=③
所以△H3=-△H1+3/2△H2=-824.35KJ/mol 该反应为放热反应,1molFe和3/2molO2反应放出824.35KJ热量.
1)中一式和二式相加消去后得H20(g)=H2O(l),ΔH=--44.即1molH20(18g)由气态到液态放热44kJ,所以1gH20由液态变为气态需要吸热44/18=2.44kJ.
(1) H2O(g)====H2(g) + ½O2(g) △H1=+241.8kJ/mol ①
H2(g) + ½O2(g)====H2O(l ) △H2=-285.8kJ/mol ②
H2O(g)====H2O(l ) △H3 ...
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(1) H2O(g)====H2(g) + ½O2(g) △H1=+241.8kJ/mol ①
H2(g) + ½O2(g)====H2O(l ) △H2=-285.8kJ/mol ②
H2O(g)====H2O(l ) △H3 ③
当1g液态水变为水蒸气时,其热量变化是多少?
①+②=③则 H2O(g)====H2O(l ) △H3==+241.8kJ/mol -285.8kJ/mol= -44kJ/mol
所以 H2O(l)====H2O(g ) △H4=44kJ/mol 因此1g水=1/18mol 所以1g水转化成水蒸汽需要吸收44kJ/mol*1/18mol=2.44KJ能量
(2)Fe2O3(s)+ 3/2 C(s)====3/2 CO2(g) + 2Fe(s) △H1=+234.1kJ/mol ①
C(s) + O2(g)====CO2(g) △H2=-393.5kJ/mol ②
Fe(s) + 3/2 O2(g)====Fe2O3(s) △H3=? ③
-①+3/2②=③
所以△H3=-△H1+3/2△H2=-824.35KJ/mol 该反应为放热反应,1molFe和3/2molO2反应放出824.35KJ热量。
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