Given the following thermochemical equations: | vedclass

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(A) The enthalpy of formation of $H_2SO_4$ corresponds to the reaction: $H_2 + S + 2O_2 \rightarrow H_2SO_4$.To obtain this,we add the given equations

Vedclass · Accepted Answer

$-814.4$

Vedclass · Answer

(A) The enthalpy of formation of $H_2SO_4$ corresponds to the reaction: $H_2 + S + 2O_2 \rightarrow H_2SO_4$.To obtain this,we add the given equations:$(1) S + O_2 \rightarrow SO_2 ; \Delta H_1 = -298.2 \ kJ$$(2) SO_2 + \frac{1}{2} O_2 \rightarrow SO_3 ; \Delta H_2 = -98.7 \ kJ$$(3) SO_3 + H_2O \rightarrow H_2SO_4 ; \Delta H_3 = -130.2 \ kJ$$(4) H_2 + \frac{1}{2} O_2 \rightarrow H_2O ; \Delta H_4 = -287.3 \ kJ$Summing these equations:$(S + O_2) + (SO_2 + \frac{1}{2} O_2) + (SO_3 + H_2O) + (H_2 + \frac{1}{2} O_2) \rightarrow SO_2 + SO_3 + H_2SO_4 + H_2O$Canceling common terms on both sides:$S + H_2 + (1 + \frac{1}{2} + \frac{1}{2}) O_2 \rightarrow H_2SO_4$$S + H_2 + 2O_2 \rightarrow H_2SO_4$The total enthalpy change is $\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 + \Delta H_4$$\Delta H = (-298.2) + (-98.7) + (-130.2) + (-287.3) = -814.4 \ kJ$.

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