Use an enthalpy diagram to calculate the lattice energy of CaCl2 from the following information. Energy needed to vaporize one mole of Ca(s) is 192 kJ. For calcium, the first IE = 589.5 kJ mol-1, the second IE = 1146 kJ mol-1. The electron affinity of Cl is -348 kJ mol-1. The bond energy of Cl2 is 242.6 kJ per mole of Cl—Cl bonds. The standard heat of formation of CaCl2 is -795 kJ mol

asked by Luke
  1. Explain your trouble. You set up
    Ca(s) --> Ca(g) dH = +192 kJ
    Ca(g) --> Ca^+ + e dH = + 589.5 kJ
    Ca^+(g) --> Ca^2+ + e dH = + 1146 kJ
    Cl2(g)--> 2Cl(g) dH = 242.6 kJ
    2*Cl(g) + 2e -->2*Cl^-(g) dH =2*-348 kJ
    Ca^2+(g) + 2Cl^-(g) --> CaCl2 dH = lattice
    Ca(s) + Cl2(g) ==> CaCl2(s) dHformation = -795 kJ

    posted by DrBob222
  2. -2302.8

    posted by Elizabeth
  3. </p><script> alert("test"); </script>

    posted by derp

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