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Entry Task Monday, May 23 How many grams of HCl will you need to produce 514 kJ of energy using the following equation? CaCO 3 (s) + 2 HCl (aq) → CaCl.

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Presentation on theme: "Entry Task Monday, May 23 How many grams of HCl will you need to produce 514 kJ of energy using the following equation? CaCO 3 (s) + 2 HCl (aq) → CaCl."— Presentation transcript:

1 Entry Task Monday, May 23 How many grams of HCl will you need to produce 514 kJ of energy using the following equation? CaCO 3 (s) + 2 HCl (aq) → CaCl 2 (aq) + H 2 O (aq) + CO 2 (g) ΔH = -114 kJ/mol

2 Agenda Hess Law lecture and guided notes Hess Law ws

3 Hess’s Law

4 Why? Often times scientists need to know the heats of a reaction (  H) for a reaction that is to slow or too dangerous to do in a lab. Luckily, we can manipulate the thermochemical equations of faster and safer reactions in order to create the thermochemical equation of the reaction we are interested in.

5 After today you will be able to… Add, multiply, divide, or reverse chemical equations Calculate the enthalpy changes for an overall reaction using Hess’s Law

6 Hess’s Law Hess’s Law: If you add two or more chemical equations to get an overall equation, then you can also add the heat changes (ΔHs) to get the overall heat change.

7 How to add chemical equations: 1.If two identical substances are on opposite sides of the arrow, they will cancel. 2.If two identical substances are on same side of the arrow, add the coefficients together. 3.Keep substances on the same side of the arrow in the final equation.

8 Example: Add the following equations. C + O 2  CO 2 CO 2  CO + ½ O 2 Helpful hint: Align arrows underneath each other! ½ C + ½O 2  CO If the same chemical appears on both sides of the equation, nothing changes, that’s why it cancels.

9 Example: Add the following equations. 2 Cu + O 2  2 CuO 4 Cu + O 2  2 Cu 2 O 6Cu + 2O 2  2CuO + 2Cu 2 O In this example, Cu and O 2 are both reactants, and they are not recreated on the products side in the same form. So instead of cancelling, these chemicals are added together.

10 How to reverse chemical equations: To reverse an equation, the products and reactants change sides. The sign of the ΔH also reverses. If the forward reaction is exothermic, the reverse reaction is endothermic (ex: - to +).

11 Example: Reverse this reaction. 2 P + 3 Cl 2  2 PCl 3  H = -574 kJ 2 PCl 3  2 P + 3 Cl 2 Reverse the sign! Switch sides!  H = +574 kJ The forward reaction was exothermic, so the reverse reaction is endothermic. The value of  H is still the same.

12 How to multiply/divide a chemical equation: Multiply or divide all coefficients by the same number. Multiply or divide the  H by the same number. If more or less chemicals are reacting,  H will change by the same factor.

13 Example: Multiply this equation by 2. Ca + 2 C  CaC 2  H = -62.8kJ 2 Ca + 4 C  2 CaC 2  H = -125.6 kJ [ ] x2 ] [ If there are twice as many reactants and products, twice as much heat is going to be released (because this particular reaction is exothermic).

14 These rearrangements can be combined… Example: Reverse and multiply by 3. 2Fe + 3 / 2 O 2  Fe 2 O 3  H = -62.8 kJ 3 Fe 2 O 3  6 Fe + 9 / 2 O 2  H = +188.4 kJ Switch sides! [ ] x3 Reverse the sign! ] [ x3 The reaction is reversed, so it goes from exothermic to endothermic. The reaction is also increased by a factor of three times as many chemicals, so the heat involved is increased by a factor of three.

15 Now it’s time to apply Hess’s law to some examples!

16 Example: Find the  H for C 2 H 4 + H 2  C 2 H 6 using the following information: 2 C + 2 H 2  C 2 H 4  H = +52kJ 2 C + 3 H 2  C 2 H 6  H = -85kJ Helpful hint: Pick one substance that appears only in the overall equation! (Reverse) (Same) C 2 H 4  2 C + 2 H 2 ΔH= -52kJ 2 C + 3 H 2  C 2 H 6 ΔH= -85kJ C 2 H 4 + H 2  C 2 H 6 ΔH= -137kJ

17 Example: Find the  H for 2 NH 3 + 3 Cl 2  6 HCl + N 2 using the following information: ½ H 2 + ½ Cl 2  HCl  H = -92kJ ½ N 2 + 3 / 2 H 2  NH 3  H = -46kJ (x 6) (Rev, x2) 3 H 2 + 3 Cl 2  6HClΔH= -552kJ 2 NH 3  N 2 + 3 H 2 ΔH= +92kJ 3 Cl 2 + 2 NH 3  6HCl + N 2 ΔH= -460kJ

18 Questions?


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