AS 2.7 Thermochemistry and Equilibrium

1. AS 2.7 Thermochemistry and Equilibrium

2. Energetic Reactions Reactions that produce heat are releasing energy. The products contain less energy than the reactants. These reactions are called exothermic reactions (the heat exits the reaction) Reactions that need energy or heat to proceed are called endothermic (the heat enters the reaction)

3. Enthalpy A Change in heat energy in reactions is called Enthalpy. A change in enthalpy is represented by ΔH (The Δ is called delta and means “change in”) To calculate changes in enthalpy we use the equation ΔH(reaction) = energy of the products – energy of reactants Exothermic reactions have negative ΔH Endothermic reactions have positive ΔH

4. Energy graphs The amount of energy in the products and reactants can be graphed.

5. When writing the equations the change in enthalpy in kJ mol-1 is included after the equation CH4(g) + H2O(g)  CO(g) + 3H2(g) ΔH =+206kJ mol-1 The enthalpy change is important in industry where the exothermic reactions can be used to power endothermic ones involved in production.

6. Calculating Enthalpy Changes Enthalpy change refers to heat energy. We can use temperature changes to calculate enthalpy changes. To do this we use the specific heat of water, s, s= 4.18J g-1 ⁰C-1 . This means it takes 4.18 Joules of energy to heat 1 gram of water 1 degree. Dilute solutions of acid or similar have the same specific heat as water.

7. Example Magnesium ribbon reacts with 2.0mL dilute hydrochloric acid. The temperature is raised by 12⁰C. Energy change is represented by ∆E Energy change = mass x temperature change x specific heat ∆E = m x ∆T x s Energy change = 2g x 12⁰C x 4.18J Energy change = 100J

8. Calorimetry: Finding Enthalpy change (∆H) • Calculate the temperature change • Calculate the energy absorbed or released. Convert it to kJ • Divide the energy change by the number of moles of reacting substances to find the enthalpy change in kJ per mol • Decide on the sign for the ∆H and write the thermochemical equation for the reaction

9. Example When 50.mL of 2.0mol L-1 sodium hydroxide solution neutralised 50.0mL of 2.0mol L-1 hydrochloric acid solution the temperature of the solution changed from 21⁰C to 35⁰C. Calculate the ∆H of the reaction assuming 1.0mL of the combined solution required 4.2J of energy to raise its temperature by 1⁰C NaOH(aq) + HCl(aq)  NaCl(aq) + H20(l)

10. When 50.mL of 2.0mol L-1 sodium hydroxide solution neutralised 50.0mL of 2.0mol L-1 hydrochloric acid solution the temperature of the solution changed from 21⁰C to 35⁰C. Calculate the ∆H of the reaction assuming 1.0mL of the combined solution required 4.2J of energy to raise its temperature by 1⁰CNaOH(aq) + HCl(aq)  NaCl(aq) + H20(l) • Calculate the temperature change Temperature change = 35⁰C - 21⁰C. = 14⁰C Calculate the energy change m(water) = 100g, ∆T= 14⁰C, s = 4.2J g-1 ⁰C-1 ∆E = m(water) x ∆T x s ∆E = 100gx 14⁰C x 4.2J g-1 ⁰C-1 ∆E = 5880 J ∆E = 5.88 kJ (3 sig. Fig)

11. When 50.mL of 2.0mol L-1 sodium hydroxide solution neutralised 50.0mL of 2.0mol L-1 hydrochloric acid solution the temperature of the solution changed from 21⁰C to 35⁰C. Calculate the ∆H of the reaction assuming 1.0mL of the combined solution required 4.2J of energy to raise its temperature by 1⁰CNaOH(aq) + HCl(aq)  NaCl(aq) + H20(l) • Calculate the enthalpy change in kJ mol-1c V(NaOH) = 50.0mL c(NaOH) = 2.0mol.L-1n(NaOH) = ? n(NaOH) = c V = 2.0mol L-1 x 50.0 x10-3 L = 0.1mol ∆H(reaction) = ∆E/n = 5.88kJ/0.1mol = 58.8kJ mol-1

12. When 50.mL of 2.0mol L-1 sodium hydroxide solution neutralised 50.0mL of 2.0mol L-1 hydrochloric acid solution the temperature of the solution changed from 21⁰C to 35⁰C. Calculate the ∆H of the reaction assuming 1.0mL of the combined solution required 4.2J of energy to raise its temperature by 1⁰CNaOH(aq) + HCl(aq)  NaCl(aq) + H20(l) 4. Decide on the sign for the ∆H and write the thermochemical equation for the reaction The temperature increased so the ∆H is negative. NaOH(aq) + HCl(aq)  NaCl(aq) + H20(l) ∆H = -58.8kJ mol.-1