Your SlideShare is downloading. ×
0
Enthalpy change
Enthalpy change
Enthalpy change
Enthalpy change
Enthalpy change
Enthalpy change
Enthalpy change
Enthalpy change
Enthalpy change
Enthalpy change
Upcoming SlideShare
Loading in...5
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×
Saving this for later? Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime – even offline.
Text the download link to your phone
Standard text messaging rates apply

Enthalpy change

41,896

Published on

Published in: Education
0 Comments
1 Like
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total Views
41,896
On Slideshare
0
From Embeds
0
Number of Embeds
1
Actions
Shares
0
Downloads
169
Comments
0
Likes
1
Embeds 0
No embeds

Report content
Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
No notes for slide

Transcript

  • 1. Enthalpy Change(ΔH)<br />
  • 2. Enthalpy Change<br />Key Concepts<br /><ul><li>The heat content of a chemical reaction is called the enthalpy (symbol: H)
  • 3. The enthalpy change (ΔH) is the amount of heat released or absorbed when a chemical reaction occurs at constant pressure.
  • 4. ΔH = H(products) - H(reactants)
  • 5. The units are usually given as kJ mol-1 (kJ/mol) or sometimes as kcal mol-1 (kcal/mol)1 calorie (1 cal) = 4.184 joules (4.184 J)
  • 6. Standard enthalpy change(ΔHѲ);Energy changes for any reaction which are measured under standard conditions of temperature and pressure25oC (298K) - 101.3 kPa (1 atmosphere) - 1 mol dm-3</li></li></ul><li>Manipulating the Enthalpy Change Term<br />N2(g) + 3H2(g) -----&gt; 2NH3(g)       ΔH = - 92.4 kJ/mol <br /><ul><li> 92.4 kJ of energy is released for every 1 mole of N2(g)
  • 7. 92.4 kJ of energy is released for every 3 moles of H2(g)
  • 8. 92.4 kJ of energy is released for every 2 moles of NH3(g) produced. </li></ul>How much energy is released if only 1 mole of ammonia (NH3) gas is produced? <br />
  • 9. a) Standard Enthalpy of formation (∆Hf0) It is the enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions of temperature and pressureEg: C(s) + O2 (g) -&gt; CO2 (g) ∆Hf0 =-393.5 KJ practice question N2(g) + 3H2(g) -----&gt; 2NH3(g)       ΔH0 = - 92.4 kJ/molFind enthalpy of formation ∆Hf0of NH3 from the equation (remember enthalpy of formation ∆Hf0 is always for 1 mole)∆Hf0 of a free element in its standard state is taken as zero.Eg: ∆Hf0 = 0 for H2(g) , N2 (g), O2(g) , F2(g), Cl2(g), Br2(l), I2 (s) , C(s,graphite)How to find ∆H0 from std. enthalpy of formation ∆H0 = Σ ∆Hf0 (products) − Σ ∆Hf0 (reactants)<br />
  • 10. Calculate enthalpy change for the reactionCH4 +2O2(g) -&gt;CO2(g) +2H2O(g) ∆Hf0of CH4 , CO2(g) , H2O(g, are -74.8, -393.5 and -285.8 kjmol-1 respectively<br />
  • 11.
  • 12.
  • 13. Enthalpy of combustion (∆Hc0) : It is enthalpy change when one mole of a substance is completely burnt in oxygen under standard conditions of temperature and pressure Eg: CH4 (g) +2O2 (g) -&gt;CO2 (g) + 2H2O (g) ∆H = – 890.3 KJ Enthalpy of solution (∆H0soln): The enthalpy change when one mole of substance is dissolved in water.Eg: NaOH (s) + H2O -----&gt; NaOH(aq) ∆Hsoln = -441.1KJ/mol Enthalpy of neutralization : The enthalpy change when one mole of acid (H+) is neutralized with one mole of base(OH-) to form one mole of water Eg: HCl (aq) + NaOH (aq) -&gt; NaCl(aq) + H2O (l) + Heat H+ (aq) + OH- (aq) -&gt; H2O (l) ∆H0 = –57.1 KJ/molThe enthalpy of neutralization of all strong acids by strong bases is always equal to -57.1 KJ .Practice Problem:<br />
  • 14. Enthalpy of neutralization<br />To find the enthalpy of neutralization for acid base reactions from experimental data. we use this rule.<br />Heat (J) = mass of solution(g) x Specific heat capacity (JK-1g-1) x change in temperature (K or C)<br /> Q = m x c x ∆T<br />Questions<br />50.0 cm3 of 1.00 mol dm-3 HCl solution was added to 50.0 cm3 of 1.00 mol dm-3 sodium hydroxide solution in a polystyrene beaker. The initial temperature of both solutions was 16.7 °C. after stirring the highest temperature reach was 23.5 °C. calculate the enthalpy change for this reaction. ( Specific heat capacity of water is 4.2JK-1g-1) <br />Answer: -57.1KJ mol-1<br />2. Chemistry book page 309 questions 1- 2<br />
  • 15. Practice problem A student used simple calorimeter to determine the enthalpy change for the combustion of ethanolC2H5OH + 3O2 -&gt; 2CO2 +3H2Owhen 0.690g of ethanol was burnt, it produced a temperature rise of 13.2K in 250g of water. Calculate ∆H for the reaction ( Specific heat capacity of water is 4.2JK-1g-1) <br />

×