Calculate the standard internal energy charge for the reaction; $$OF2(g)+H2O(g)$$→$$O2(g)+2HF(g)at$$ $$298$$ K. The standard enthalpies of formation of $$OF2(g)$$, $$H2O(g)$$,$$HF(g)$$ are $$+20$$,$$+250$$ , and −$$270kJmol$$−$$1$$ respectivley?

Question

Calculate the standard internal energy charge for the reaction; $$OF2(g)+H2O(g)$$→$$O2(g)+2HF(g)at$$ $$298$$ K. The standard enthalpies of formation of $$OF2(g)$$, $$H2O(g)$$,$$HF(g)$$ are $$+20$$,$$+250$$ , and −$$270kJmol$$−$$1$$ respectivley?

Infinity Learn · Accepted Answer

ΔHReaction ∘$$=$$∑HProducts ∘−∑HReactants ∘$$=2$$×HHF∘(g)+H$$∘$$O2(g)−H∘$$OF2(g)+HH2O$$∘(g)∴H∘ of free elements $$=0$$∴$$HO2$$∘$$=0$$∴ΔHf∘$$=H$$∘, i.e., standard heat of formation $$=$$ standard heat enthalpy of a compoundAlso, ΔHf∘ for $$F2O$$ is $$+ve$$;ΔHf∘ for $$H2O$$ is $$-ve$$,  because heat of combustions are exothermic.ΔHR∘$$=$$[$$2$$×$$($$−$$270)$$]$$+0$$−[$$20+($$−$$250)$$]$$=$$−$$310kJ$$ Now ΔH∘$$=$$ΔU∘$$+$$ΔnRT∴ΔHR∘$$=$$−$$310$$×$$103J$$;Δ$$n=3$$−$$2=1R=8.314J$$;$$T=298K$$∴ −$$310$$×$$103=$$ΔU∘$$+1$$×$$8.314$$×$$298$$∴ ΔU∘$$=$$−$$312477.5joule$$∴ ΔU∘$$=$$−$$312.4775kJ$$

Calculate the standard internal energy charge for the reaction; OF2(g)+H2O(g)→O2(g)+2HF(g)at 298 K. The standard enthalpies of formation of OF2(g), H2O(g),HF(g) are +20,+250 , and −270kJmol−1 respectivley?

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