CO(g) + H2O(g) → CO2(g) + H2(g)
From the reaction above, calculate the standard enthalpies of formation if CO2(g), H2O(g) and CO(g) in kJ mol-1 are -394, -242, and -110 respectively.
-262 kJ mol-1
-42 kJ mol-1
+42 kJ mol-1
+262 kJ mol-1
Explanation
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Discussions (12)
First of all according to your calculations you said that -242+110 is -352 which is absolutely wrong the question all one needs to do is input the given values in their respective slot in the equation I.e co is -110,H2O is -242 and co2 is -394 so inputting this values we have :-110-242=-394+ H2 so we have -352=-394+H2 = -352+394=H2
42=H2 and since it has no sigh behind it that means it is a positive value.
I saw this question some where and the ans there was +42... ....Guys! i think its normal solving.or is there something else? Am freaked out oooo.....
it correctly.
∆H= {£P} - {£R}
{CO2}-{CO+H20}
{-394+0} -{-110-242}
{-394} - {-352}
-394+352
∆H= -42
