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Given the following data : Determin...

Given the following data :

Determine at what temperature the following reaction is spontanous?
`FeO(s)+C_("graphite")toFe(s)+CO(g)`

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Given the following data: Determine at what temperature the following reaction is spontaneous? Fe O (s) + C (Graphite) to Fe (s) + CO (g)

Given the following data {:("Substance",DeltaH^(@)(KJ//mol),S^(@)(J//mol K),DeltaG^(@)(kJ//mol),),(FeO(s),-266.3,57.49,-245.12,),(C("Graphite"),0,5.74,0,),(Fe(s),0,5.74,0,),(CO(g),-110.5,197.6,-137.15,):} Determine at what temperature the following reaction is spontaneous? FeO(s)+C("Graphite")rarrFe(s)+CO(g)

Given the following data {:("Substance",DeltaH^(@)(KJ//mol),S^(@)(J//mol K),DeltaG^(@)(kJ//mol),),(FeO(s),-266.3,57.49,-245.12,),(C("Graphite"),0,5.74,0,),(Fe(s),0,5.74,0,),(CO(g),-110.5,197.6,-137.15,):} Determine at what temperature the following reaction is spontaneous? FeO(s)+C("Graphite")rarrFe(s)+CO(g)

Given the following data: {:("Substance", DeltaH^(@)"(KJ//mol)", S^(@)"(J//mol K)", DeltaG^(@)"(KJ//mol)"),(FeO(s),-266.3,57.49,-245.12),(C("Graphite"), 0, 5.74, 0),(Fe(s),0,27.28,0),(CO(g),-110.5,197.6,-137.15):} Determine at what temperature the following reaction is spontaneous? " "FeO(S) + C("Graphite") rarr Fe(s) + CO(g)

Determine DeltaS_(Total) and decide whether the following reaction is spontaneous at 298 K. Fe_2O_(3(s))+ 3CO_((g)) rarr 2Fe_(s) + 3CO_(2(g)) DeltaH^@ =- 24.8 kj, DeltaS^@ = + 15 JK^(-1)

At 1250 K 2FeO(s)to2Fe(s)+O_(2)(g) , DeltaG=+320kJmol^(-1) of O_(2) 2C(s)+O_(2)(g)to2CO(g) , DeltaG=-430 kJ mol^(-1) of O_(2) The following coupled reaction is spontaneous because 2FeO(s)+2C(s)to2Fe(s)+2CO(g)

At 1250 K 2FeO(s)to2Fe(s)+O_(2)(g) , DeltaG=+320kJmol^(-1) of O_(2) 2C(s)+O_(2)(g)to2CO(g) , DeltaG=-430 kJ mol^(-1) of O_(2) The following coupled reaction is spontaneous because 2FeO(s)+2C(s)to2Fe(s)+2CO(g)

Calculate the temperature above which the reaction of lead oxide to lead in the following reaction become spontaneous. PbO(s) + C(s) rarr Pb(s) + CO(g) Given Delta H = 108.4 kJ mol^(-1), Delta S = 190 JK^(-1) mol^(-1)