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Solutions Form 0

# Solutions Form 0 - Solutions Form 0 1 Solution This is a...

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Unformatted text preview: Solutions Form 0 1 Solution: This is a Hess Law problem. The given reactions are added and subtracted to produce the desired reaction. Ask yourself how can I use these reactions to make B 2 H 6 ¡ g on the reactants side? (Always start with a species that exists in only one of the given equations. So starting with O 2 would be a bad idea.) To make B 2 H 6 ¡ g on the reactants side use equation 3 as is. Then how can I make a 3H 2 O ¡ l on the products side? Reaction 2 times 3 2 . Let¡s see what we have now. RXN 3. B 2 H 6 ¡ g v 2B ¡ s ¡ 3H 2 ¡ g H ( ¢ " 35.4 kJ RXN 2. 3 2 (2 H 2 ¡ g ¡ O 2 ¡ g v 2 H 2 O ¡ l H ( ¢ " 571.7 kJ) SUM B 2 H 6 ¡ g ¡ 3 2 O 2 ¡ g v 2B ¡ s ¡ 3H 2 O ¡ l H ( ¢ " 35.4 kJ ¡ 3 2 ¡ " 571.7kJ Notice that the 3H 2 ¡ g on both sides of the equation cancels. Now we need a way to make B 2 O 3 ¡ s on the products side. Do that with 1 2 rxn 1. I haven¡t thought about the O 2 and B ¡ s . They occur in more than 1 equation. I am just hoping that if I get everything else right those ingredients will work themselves out. SUM above B 2 H 6 ¡ g ¡ 3 2 O 2 ¡ g v 2B ¡ s ¡ 3H 2 O ¡ l H ( ¢ " 35.4 kJ ¡ 3 2 ¡ " 571.7kJ RXN 1. 1 2 (4B ¡ s ¡ 3O 2 ¡ g v 2B 2 O 3 ¡ s H ( ¢ " 2509.1 kJ) B 2 H 6 ¡ g ¡ 3O 2 ¡ g v B 2 O 3 ¡ s ¡ 3H 2 O ¡ l H ( ¢ " 35.4 kJ ¡ 3 2 " 571.7 kJ ¡ 1 2 " 2509.1 kJ ¢ " 2147.5 kJ 2 Solution: Choice A and E are parts of the second law of thermodynamics. The first law says E ¢ q ¡ w which is a statement of the conservation of energy. 3 Solution: Sometimes a question like this is given by describing the curve in words rather than by using a figure. Close your eyes and ask yourself if someone described this curve to me in words could I still identify the various energies G,K,H,J? Read over the notes in Chapter 14 if you still don¡t get this. 4 Solution: For any general reaction a A ¡ b B v c C ¡ d D The equilibrium constant is K ¢ ¢ C £ c ¢ D £ d ¢ A £ a ¢ B £ b Common Mistakes: Remember that the concentrations of pure liquids and solids is 1. The ingredients like Fe ¡ s and Fe 2 O 3 ¡ s do not appear in the equilibrium expression. Get into the habit of crossing them out in the chemical equation so you don¡t forget when you build the equilibrium expression. Don¡t forget to use the stoichiometry in the exponent. Remember it¡s K ¡ products stoichiometry reactants stoichiometry 5 Solution: The oxidation states are 3 ¢ 2 Ni 2 ¢ ¡ aq ¢ 2 Al ¡ s v 3 Ni ¡ s ¢ 2 3 ¢ Al 3 ¢ ¡ aq The half cell reactions are reduction Ni 2 ¢ ¡ aq ¢ 2 e " v Ni ¡ s cathode oxidation Al ¡ s v Al 3 ¢ ¢ 3 e " anode Electrons always flow from the anode to the cathode in the spontaneous cell. This cell is spontaneous since E cell ( ¡ ¢ 1.39 V. The anode is always where oxidation occurs....
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Solutions Form 0 - Solutions Form 0 1 Solution This is a...

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