Question

Baking soda decomposes on heating as follows, creating the holes in baked bread: 2NaHCO3(s) — Na,CO() + CO2(g) +H,0(1) AH --129.3 kJ 1st attempt Il See Periodic Table O See Hint Calculate the standard molar enthalpy of formation of NaHCO3(s) from the enthalpy of the baking soda decomposition reaction and the following information: AH9 [Na2CO3(s)] - -1131kJ AHCO2(g)] = -394 kJ AH9 [H2000) = -286 kJ Standard molar enthalpy:

Answer 1

Here the reactants are 2 moles of NaHCO₃

The products are 1 mole of Na₂CO_3, H₂O and CO₂ each.

Using the variables as mentioned in the question, The formula for heat of reaction is,

$\Delta H_{rxn}= \sum \Delta H_f(Products)- \sum \Delta H_f(Reactants)\\ \Rightarrow \Delta H_{rxn}=[\Delta H_f(Na_2CO_3(s))+\Delta H_f(CO_2(g))+\Delta H_f(H_2O(l))]- 2\Delta H_f(NaHCO_3(s))\\ \Rightarrow -129.3 kJ=[(-1131kJ)+(-394kJ)+(-286kJ)]-2\Delta H_f(NaHCO_3(s))\\ \Rightarrow 2\Delta H_f(NaHCO_3(s))=-1681.7kJ\\ \Rightarrow \Delta H_f(NaHCO_3(s))=-840.85kJ\\$

Thus, the answer will be -841 kJ (Upto 3 significant figures)