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5 Thermochemistry Solutions to Exercises (d) The temperature of the calorimeter rises, so the reaction is exothermic and the sign of q is negative. The reaction as carried out involves only 0.050 mol of CuSO₄ and the stoichiometrically equivalent amount of KOH. On a molar basis, for the reaction as written in part (c) 0.050 5.116 (a) NaNO₃(aq) + AgCl(s) net ionic equation: (aq) + AgCl(s) (b) for the complete molecular equation will be the same as for the net ionic equation. Na (aq) and (aq) are spectator ions; they appear on both sides of the chemical equation. Since the overall enthalpy change is the enthalpy of the products minus the enthalpy of the reactants, the contributions of the spectator ions cancel. (c) = NaNO₃(aq) + AgCl(s) AgNO₃(aq) - NaCl(aq) AgCl(s) 5.117 (a) 21.83 X 1 mol X mol 1 mol C mol = g C The sample mass is (5.9572 + 0.5001) = 6.457 (b) 5.957 = 0.4960 C; 0.4960/0.496 = 1 0.496/0.496 The empirical formula of the hydrocarbon is CH. (c) Calculate for 6.457 g of the sample. 21.83 + 4.47 H₂O(g), = -311 kJ 137