Question

Question 5 (4 points) What is the standard molar heat of solution for solid calcium bromide (CaBr2) given the standard enthalpy of formation data below? CaBrz(s) → Ca2+ (aq) + 2 Br(aq) AHºf of CaBrz(s) is -682.8 kJ.mol-1 AH° of Ca2+(aq) is -542.83 kJ.mol-1 AH, Of Br(aq) is -121.55 kJ mol-1 - 1347.2 kJ.mol-1 -103.1 kJ-mol-1 -21.6 kJ-mol-1 +21.6 kJ-mol-1

Answer 1

Q5. Enthalpy of reaction = sum of enthalpies of products - sum of enthalpies of reactant

For the reaction

CaBr_{2 (s)}$\rightarrow$ Ca²⁺_(aq)+ 2Br^-(aq)

AH reaction = AHCalaq + 2 x ΔΗΨBraq) - ΔΗCaBr2(s)

= -542.83 + 2x(-121.55) - ( - 682.8) = -103.13 kJ/mole

Q6.

V= 10L, P = 1.13atm, T = 100⁰ C = 373 K,

Substituting in PV = nRT

1.13atm x 10L = n x 0.0821(L.atm)/ (K.mol)x 373 K

n = (1.13x10)/ (0.0821x373) = 0.3690 ( all units cancels out)

n = no of moles = Weight/ Mol Wt

or 10 g/Mol.wt = 0.3690

or Mol wt = 10/ 0.3690 = 27g/mol

so the gas is HCN whose molecular weight is 27 g (1+12+14)