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/mo |
D.2H2 ( g ) + O2 ( g ) = 2H2O ( g )△H =" +484" kJ/mol |