Calculate ΔHrxn for the following reaction:
CaO(s)+CO2(g)→CaCO3(s)
Use the following reactions and given ΔH′s.
Ca(s)+CO2(g)+12O2(g)→CaCO3(s), ΔH= -812.8 kJ
2Ca(s)+O2(g)→2CaO(s), ΔH= -1269.8 kJ
Ca + CO2 +12O2→ CaCO3,..... ΔH= -812.8 kJ........(1)
2Ca + O2 → 2CaO(s), ............ΔH= -1269.8 kJ.........(2)
multiplying (1) by 2
2Ca + 2CO2 + O2 -------> 2CaCO3 .....ΔH = -812.8 X 2 = -1625.6 kj .......(3)
subtracting (2) from (3)......
2Ca + 2CO2 + O2 - 2Ca - O2 --------> 2CaCO3 - 2CaO ......ΔH = -1625.6 - -1269.8 = -1625.6 + 1269.8 = -355.8 kJ
2CO2 --------> 2CaCO3 - 2CaO .....ΔH = -355.8 kJ
2CO2 + 2CaO --------> 2CaCO3 ....ΔH = -355.8 kJ
dividin by 2
CO2 + CaO --------> CaCO3 .......ΔH = -177.9 kJ
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Verified answer
Ca + CO2 +12O2→ CaCO3,..... ΔH= -812.8 kJ........(1)
2Ca + O2 → 2CaO(s), ............ΔH= -1269.8 kJ.........(2)
multiplying (1) by 2
2Ca + 2CO2 + O2 -------> 2CaCO3 .....ΔH = -812.8 X 2 = -1625.6 kj .......(3)
subtracting (2) from (3)......
2Ca + 2CO2 + O2 - 2Ca - O2 --------> 2CaCO3 - 2CaO ......ΔH = -1625.6 - -1269.8 = -1625.6 + 1269.8 = -355.8 kJ
2CO2 --------> 2CaCO3 - 2CaO .....ΔH = -355.8 kJ
2CO2 + 2CaO --------> 2CaCO3 ....ΔH = -355.8 kJ
dividin by 2
CO2 + CaO --------> CaCO3 .......ΔH = -177.9 kJ