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Do Now Day 6

Do Now Day 6. Draw what the heating curve would look like for a sample of liquid water at 35.0°C heated to steam at 120.0°C. Use the graph above to calculate the total energy required to raise the temperature of 225 g of water from 35.0°C to 120.0°C. (Hint: Only 3 calculations are required!).

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Do Now Day 6

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  1. Do NowDay 6 • Draw what the heating curve would look like for a sample of liquid water at 35.0°C heated to steam at 120.0°C. • Use the graph above to calculate the total energy required to raise the temperature of 225 g of water from 35.0°C to 120.0°C. (Hint: Only 3 calculations are required!)

  2. Do NowDay 6 • Draw what the heating curve would look like for a sample of liquid water at 35.0°C heated to steam at 120.0°C. 100˚C q=mcΔT 120˚C q=mHv 35˚C q=mcΔT 0˚C Temp (˚C) Time

  3. Do NowDay 6 2. Use the graph above to calculate the total energy required to raise the temperature of 225 g of water from 35.0°C to 120.0°C. (Hint: Only 3 calculations are required!) (1) Liquid, q=mcΔT (3) Gas, q=mcΔT q=(225g)(4.18J/g˚C)(100˚C - 35˚C) q=(225g)(2.02 J/g˚C)(120˚C - 100˚C) q=(225g)(4.18J/g˚C)(65˚C) q=(225g)(2.02J/g˚C)(20˚C) q=+61,132.5J q=+9,090 J (2) Boiling, q=mHv 61,132.5 508,500 q=(225g)(2260 J/g)) + 9,090 q=+508,500 J 578,722.5 J

  4. Day 6 - Notes Unit: Thermochemistry Hess’s Law

  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: • If two identicalsubstances are on opposite sides of the arrow, they will cancel (reduce). • If two identicalsubstances are on same side of the arrow, add the coefficients together. • Keep substances on the same side of the arrow in the final equation.

  8. Helpful hint: Align arrows underneath each other! Example: Add the following equations. C + O2 CO2 CO2 + 2H2O  CH4 + 2O2 C + O2 + 2 H2O CH4 + 2O2

  9. Example: Add the following equations. 2 CO + O2 2 CO2 C + O2 CO2 2CO + C+ 2O2 3CO2

  10. How to reverse chemical equations: • To reverse an equation, change sides with the products and reactants, and change the sign on the ΔH (ex: - to +)

  11. Reverse the sign! Example: Reverse this reaction. H2 + I2 2 HI H = +53.0 kJ Switch sides! 2 HI  H2 + I2 H = -53.0 kJ

  12. How to multiply/divide a chemical equation: • Multiply or divide the coefficients by the same number and also multiply/ divide the H by the same number.

  13. Example: Multiply this equation by 2. 4 Al+ 3 O2 2 Al2O3H = -3351kJ [ ] [ ] x2 x2 8 Al+ 6 O2 4 Al2O3 H = -6702 kJ

  14. These rearrangements can be combined… Example: Reverse and multiply by 3. 2C + 3H2 C2O6H = -84.0 kJ [ ] [ ] x3 x3 Reverse the sign! Switch sides! 3 C2H6 6 C + 9 H2 H = +252.0 kJ

  15. It’s time to apply Hess’s law to some examples!

  16. Helpful hint: Pick one substance that appears only in the overall equation! (Reverse) Example: Find the H for C2H4 + H2 C2H6 using the following information: 2 C + 2 H2 C2H4H = +52. 4kJ 2 C + 3 H2 C2H6H = -84.0 kJ (Same) C2H4 2 C + 2 H2 ΔH= -52kJ 2 C+ 3 H2 C2H6 ΔH= -85kJ C2H4 + H2  C2H6 ΔH= -137kJ

  17. (x 3) Example: Find the H for 2 NH3 + 3 Cl2 6 HCl + N2 using the following information: H2 + Cl2 2 HCl H = -184 kJ N2 + 3 H2 2NH3H = -92 kJ (Rev) 3 H2 + 3 Cl2 6HCl ΔH= -552kJ 2 NH3 N2 + 3 H2 ΔH= +92kJ 3 Cl2 + 2 NH3 6HCl + N2 ΔH= -460kJ

  18. Questions? Complete WS 6

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