Given that: 2SO3 (g) ----> 2SO2 (g) + O2 (g) delta Hrxn^0 = +198.2 Calculate del

Question

Given that: 2SO3 (g) ----> 2SO2 (g) + O2 (g) delta Hrxn^0 = +198.2 Calculate delta H^0 for the following reaction: SO2 (g) + 1/2 O2 (g) -----> SO3 (g) please show all steps

Explanation / Answer

1. given : A) S & O2 -- SO2 dH = -297kJ B) 2SO2 (g) + O2 (g) --> 2SO3 (g) dH = -198 kJ find S & 1.5 O2 --> SO3 can be found by using 1 equation (A) to provide 1 Sulfur as a reactant using 1/2 of equation (B) to provide 1 SO3 as a product 1 (A) S & O2 -- SO2 dH = -297kJ 1/2 (B) 1SO2 (g) + 0.5 O2 (g) --> 1 SO3 (g) dH = -99 kJ combining all reactants --> combining all products S & 1.5 O2 & SO2 --> SO2 & SO3 SO2's cancel out... giving you the equation that you wish: S & 1.5 O2 --> SO3 Hess's law says that if 1 (A) & 1/2 (B) gives you the equation that you wish, then combining the energy fro (A) & 1/2 (B) , will give you the energy you wish -297kJ & -99 kJ = -396 kJ ======================================… Is the answer -396 kJ? That is the number in the appendix of my book for SO3, but that seems to obvious! 2.. How much energy is evolved during the reaction of 32.5 g B2H6 and 72.5 g Cl2? find moles of Cl2, using molar mass : 72.5 g Cl2 divided by 70.90g / mol = 1.023 moles of Cl2 by the equation B2H6 (g) + 6 CL2 (g) --> 2BCl3 (g) + 6 HCl (g) & 1396 kJ released 6 moles of Cl2 rreacts to release 1396 kJ so 1.023 moles of Cl2 @ 1396 kJ / 6 mol Cl2 = 238 kJ

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