Given the following equations and `DeltaH^(@)` values, determine the enthalpy of reaction at 298 K for the reaction : `C_(2)H_(4)(g)+6F_(2)(g) to 2HF_(4)(g)+4HF(g)` `H_(2)(g)+F_(2)(g)to2HF(g)+2HF(g)," "DeltaH_(1)^(@)=-537" kJ"` `C(s)+2F_(2)(g)toCF_(4)(g)," "DeltaH_(2)^(@)=-680" kJ"` `2C(s)+2H_(2)(g)toC_(2)H_(4)(g)," "DeltaH_(3)^(@)=52" kJ"`
Given the following equations and values, determine the enthalpy of reaction at 298 K for reaction: C_(2)H_(4)(g) + 6F_(2)(g) rarr 2CF_(4)(g) + 4HF(g) H_(2)(g) + F_(2)(g) rarr 2HF(g) " "Delta H_(1)^(@) = -537 kJ C(s) + 2F_(2)(g) rarr CF_(4)(g) " "DeltaH_(2)^(@) = -680 kJ 2C(s) + 2H_(2)(g) rarr C_(2)H_(4)(g) " "Delta H_(3)^(@) =52 kJ
For the reaction C_(2)H_(4)(g)+3O_(2)(g) rarr 2CO_(2) (g) +2H_(2)O(l) , Delta E=-1415 kJ . The DeltaH at 27^(@)C is
Determine Delta U^(@) at 300K for the following reaction using the listed enthalpies of reaction : 4CO(g)+8H_(2)(g)to3CH_(4)(g)+CO_(2)(g)+2H_(2)O(l) C("graphite")+(1)/(2)O_(2)(g)to CO(g), DeltaH_(1)^(@)=-110.5KJ CO(g)(1)/(2)O_(2)(g)toCO_(2)(g), DeltaH_(2)^(@)=-282.9KJ H_(2)(g)+(1)/(2)O_(2)(g)to H_(2)O(l), DeltaH_(3)^(@)=-285.8KJ C("graphite")+2H_(2)(g)to CH_(4)(g), DeltaH_(4)^(@)=-74.8KJ
Calculate in kJ for the following reaction : C(g) + O_(2)(g) rarr CO_(2)(g) Given that, H_(2)O(g) + C(g) + H_(2)(g) , Delta H = +131 kJ CO(g) + 1/2 O_(2)(g) rarr CO_(2)(g), " " Delta H = -242 kJ H_(2)(g) + 1/2 O_(2)(g) rarr H_(2)O(g), " "DeltaH = -242 kJ
Use the given standard enthalpies of formation (in kJ/mol) to determine the enthalpy of reaction of the following reaction : NH_(3)(g)+3F_(2)(g)rarrNF_(3)+3HF(g) DeltaH_(f)^(@)(NH_(3),g)=-46.2, " "DeltaH_(f)^(@)(NF_(3),g)=-113.0 ," "DeltaH_(f)^(@)(HF, g)=-269.0
For the reaction, N_(2)H_(4)(l) to N_(2)(g) + 2H_(2)(g) DeltaH^(@) =-50.6kJ. This reaction is:
NARENDRA AWASTHI-THERMODYNAMICS-Level 3 - Match The Column