Given: S + O_2 rightarrow SO_2 : Delta H_1 = | vedclass

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(A) The heat of formation of $H_2SO_4$ is the enthalpy change for the reaction: $S(s) + H_2(g) + 2O_2(g) \rightarrow H_2SO_4(l)$.By adding the given t

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$-814.4$

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(A) The heat of formation of $H_2SO_4$ is the enthalpy change for the reaction: $S(s) + H_2(g) + 2O_2(g) \rightarrow H_2SO_4(l)$.By adding the given thermochemical equations:$(1) \ S + O_2 \rightarrow SO_2, \Delta H_1 = -298.2 \ kJ$$(2) \ SO_2 + 1/2 \ O_2 \rightarrow SO_3, \Delta H_2 = -98.7 \ kJ$$(3) \ SO_3 + H_2O \rightarrow H_2SO_4, \Delta H_3 = -130.2 \ kJ$$(4) \ H_2 + 1/2 \ O_2 \rightarrow H_2O, \Delta H_4 = -287.3 \ kJ$Summing these equations:$(S + O_2) + (SO_2 + 1/2 \ O_2) + (SO_3 + H_2O) + (H_2 + 1/2 \ O_2) \rightarrow SO_2 + SO_3 + H_2SO_4 + H_2O$Canceling common terms on both sides gives:$S + H_2 + 2O_2 \rightarrow H_2SO_4$Therefore,$\Delta H_f(H_2SO_4) = \Delta H_1 + \Delta H_2 + \Delta H_3 + \Delta H_4$$\Delta H_f = -298.2 + (-98.7) + (-130.2) + (-287.3) = -814.4 \ kJ$.

Similar Questions

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