Write reactions for which the enthalpy change will be …..?

3 Answers

• the standard enthalpy is @ 25C, so I am using product in the physical state they would be @ 25C

  a)∆H f For solid aluminum oxide.

  2Al (s) & 1.5 O2 (g) –> 1 Al2O2 (s)

  (formation reactions only produce 1 mole of product)

  b) the standard enthalpy of combustion of liquid ethanol , C2H5OH .

  1 C2H5OH (l) & 3 O2 (g) –> 2CO2 (g) & 3 H2O (l)

  C)the standard enthalpy of nuetralization of sodium hydroxide solution by hydrochloric acid.

  1 NaOH (aq) & 1 HCl (aq) –> NaCl (aq) & H2O (l)

  d)∆Hf for gaseous vinyl chloride , C2H3Cl(g)

  2 C (s, graphite) & 1.5 H2 (g) & 1/2 Cl2 (g) –> 1 C2H3Cl(g)

  e) the enthalpy of combustion of liquid benzene, C6H8 (l)

  Benzene’s formula is C6H6, & balances with whole numbers as:

  2C6H6 (l) & 15 O2 (g) –> 12 CO2 (g) & 6 H2O (g)

  however the standard enthalpy of combustion of liquid benzene is for the combustion of 1 mole

  C6H6 (l) & 7.5 O2 (g) –> 6 CO2 (g) & 8 H2O (g)

  f)the enthalpy of solution of solid ammonium bromide.

  usually written as NH4Br (s) —> NH4Br (aq)

  [but sometimes writen as: NH4Br (s) —> (NH4)+1 (aq) & Br- (aq) ]

• in case you want to have a usual attitude to sparkling up this style of question, positioned the means (enthalpy replace) interior the reaction! Exothermic (adverse enthalpy replace) is going on the right area (as a product) and endothermic (functional enthalpy replace) is going on the left (as a reactant). as a outcome: CH4(g) + 2 O2(g) –> CO2(g) + 2 H2O(l) + 890kJ you could now use stoichiometry! a. 17 mol H2O * (890kJ / 2H2O) = 7565kJ b. 11 mol O2 * (890kJ / 2 O2) = 4895kJ

• aluminum solid to liquid