consider the following reaction.
H2 (g) 2CO(g) ---> H2O2(l) + 2C(s) DeltaH = +22.2kj
Calculate the enthalphy change when 5 moles of CO is consumed
2007-01-26 06:08:06 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
enthapy change for 1 mole of CO to interact : 22.2/2 = 11.1 kj
So, enthalp change for 5 moles CO is 11.1kj x 5 = 55.5 kj
2007-01-26 06:17:07 · answer #1 · answered by sleepingprince88 2 · 0⤊ 0⤋
In the given equation H denotes enthalpy and deltaH is the enthalpy change. Note that the reaction is balanced. Therefore, for 2 moles of CO consumed, the enthalpy change (i.e., deltaH) is 22.2 kj.
=> for 1 mole of CO consumed, deltaH = 22..2kj/2 = 11.1 kj
=> for 5 moles of CO consumed, deltaH = 11.1 kj *5 = 55.5kj.
2007-01-26 14:21:46 · answer #2 · answered by gaurav_dwivedi012 2 · 0⤊ 0⤋
Since equation tells you: 2 moles of CO gives 22.2KJ
then 5moles of CO gives (5/2) x 22.2 = 55.5KJ
2007-01-26 14:16:01 · answer #3 · answered by Alan S 3 · 0⤊ 0⤋
If 2 moles results in 22.2 kj, then 5 will result in 55.5.
2007-01-26 14:14:30 · answer #4 · answered by gebobs 6 · 0⤊ 0⤋
I suggest that you have the equation wrong. It should be:
H2(g) + CO(g) ===> H2O(l) + C(s) DeltaH = +22.2kJ
2007-01-26 14:34:14 · answer #5 · answered by steve_geo1 7 · 0⤊ 1⤋
that one is too hard for me
2007-01-26 14:14:37 · answer #6 · answered by kurticus1024 7 · 0⤊ 1⤋