摘要:已知在1×105 Pa.298 K条件下.2 mol氢气燃烧生成水蒸气放出484 kJ热量.下列热化学方程式正确的是 A.H2O ( g ) = H2 ( g ) + 1/2O2 ( g ) △H = +242 kJ/mol B.2H2 ( g ) + O2 ( g ) = 2H2O ( l ) △H = -484 kJ/mol C.H2 ( g ) + 1/2O2 ( g ) = H2O ( g ) △H = +242 kJ/mol D.2H2 ( g ) + O2 ( g ) = 2H2O ( g ) △H = +484 kJ/mol

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