Question

Given the following information below, use Hess’s Law to calculate the enthalpy of formation for sodium oxide:

Na (s)     +      HCl (l)  à    NaCl (aq) + ½ H₂ (g)                ∆H_Rx = -394.1 kJ/mol

Na₂O (s)     +     2 HCl (l)  à 2 NaCl (aq)   + H₂O              ∆H_Rx = -691.7 kJ/mol

H₂ (g)       +      ½ O₂ (g)  à    H₂O (g)                              ∆H_Rx = -284.3 kJ/mol

2 Na (s)       +       ½ O ₂ (g)   à   Na ₂O (s)                           ∆ H_Rx =    __________ kJ/mol

Calculated Heat of Reaction is....?

Suppose you had used 4x the amount of grams of NaOH(s) in the reaction: NaOH (s) --> Na⁺ + OH^- with an enthalpy value (H_rxn)= -46.80 kJ/mol. What affect would this have on the reaction? (choose all that apply)

a) The enthalpy value will be higher.

b) The enthalpy value will remain the same.

c) The enthalpy value will be less.

d) The enthalpy value will be less.

e) The heat released will be less.

Answer 1

According to Hess's law enthalpy of reaction is the sum of enthalpy of more then one steps .