If S + O_2 to SO_2; (Delta H = -298.2 kJ),SO | vedclass

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(D) The formation reaction of $H_2SO_4$ is: $H_2 + S + 2O_2 	o H_2SO_4$.To obtain this,we add the given equations:$(1) S + O_2 	o SO_2; \Delta H_1 =

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$-813.9$

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(D) The formation reaction of $H_2SO_4$ is: $H_2 + S + 2O_2 	o H_2SO_4$.To obtain this,we add the given equations:$(1) S + O_2 	o SO_2; \Delta H_1 = -298.2 \ kJ$$(2) SO_2 + \frac{1}{2} O_2 	o SO_3; \Delta H_2 = -98.2 \ kJ$$(3) SO_3 + H_2O 	o H_2SO_4; \Delta H_3 = -130.2 \ kJ$$(4) H_2 + \frac{1}{2} O_2 	o H_2O; \Delta H_4 = -287.3 \ kJ$Adding $(1) + (2) + (3) + (4)$:$S + O_2 + SO_2 + \frac{1}{2} O_2 + SO_3 + H_2O + H_2 + \frac{1}{2} O_2 	o SO_2 + SO_3 + H_2SO_4 + H_2O$Canceling common terms on both sides gives:$H_2 + S + 2O_2 	o H_2SO_4$$\Delta H_f = \Delta H_1 + \Delta H_2 + \Delta H_3 + \Delta H_4$$\Delta H_f = (-298.2) + (-98.2) + (-130.2) + (-287.3) = -813.9 \ kJ$.

If $S + O_2 \to SO_2; (\Delta H = -298.2 \ kJ)$,$SO_2 + \frac{1}{2} O_2 \to SO_3; (\Delta H = -98.2 \ kJ)$,$SO_3 + H_2O \to H_2SO_4; (\Delta H = -130.2 \ kJ)$,$H_2 + \frac{1}{2} O_2 \to H_2O; (\Delta H = -287.3 \ kJ)$,then the enthalpy of formation of $H_2SO_4$ at $298 \ K$ will be......$kJ$.

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