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प्रश्न
The enthalpies of elements in their standard states are taken as zero. The enthalpy of formation of a compound ______.
पर्याय
is always negative
is always positive
may be positive or negative
is never negative
उत्तर
The enthalpies of elements in their standard states are taken as zero. The enthalpy of formation of a compound may be positive or negative.
Explanation:
Heat of formation of a compound may be positive or negative, e.g.,
\[\ce{C(s) + O2(g) -> CO2(g); ΔH° = - 393.5 kJ mol^{-1}}\]
\[\ce{N2(g) + 1/2 O2(g) -> N2O(g); ΔH° = + 92 kJ mol^{-1}}\]
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संबंधित प्रश्न
Standard molar enthalpy of formation, ∆fHΘ is just a special case of enthalpy of reaction, ∆rHΘ. Is the ∆rHΘ for the following reaction same as ∆fHΘ? Give reason for your answer.
\[\ce{CaO(s) + CO2(g) -> CaCO3(s); ∆_fH^Θ = - 178.3 kJ mol^{-1}}\]
The enthalpy of reaction for the reaction: \[\ce{2H2 (g) + O2(g) -> 2H2O (l)}\] is ∆rHΘ = – 572 kJ mol–1. What will be standard enthalpy of formation of \[\ce{H2O (l)}\]?
The enthalpy change accompanying the formation of one mole of a compound from its elements, all the substance being in their standard states, is called as ______.
Enthalpy for the reaction \[\ce{Ag+ (aq) + Br- (aq) -> AgBr(s)}\] is : 90 kJ. Magnitude of enthalpy of formation of Ag+ (aq) and Br- (aq) are in the ratio 5 : 6. Formation of Ag+(aq) is an endothermic process whereas formation of Br is an exothermic process. Enthalpy of formation of AgBr is - 110 kJ/ mol. The enthalpy of formation of Ag+ (aq) will be ______ kJ/mol.
Enthalpy for the reaction \[\ce{Ag^+ (aq) + Br^- (aq) -> AgBr(s)}\] is −84.54 kJ. Magnitude of enthalpy of formation of Ag+ (aq) and Br− (aq) are in the ratio 8 : 9. Formation of Ag+ (aq) is an endothermic process whereas formation of Br− (aq) is an exothermic process. Enthalpy of formation of AgBr is −99.54 kJ/mol. The enthalpy of formation of Ag+ (aq) is ______ kJ/mol.
