Question

Baking soda decomposes on heating as follows, creating the holes in baked bread: Calculate the standard molar enthalpy of formation of NaHCO3(s) from the enthalpy of the baking soda decomposition reaction and the following information:?

2NaHCO3(s)----> Na2CO3(s)+CO2(g)+H2O(l)

ΔH°rxn = -129.3 kJ

ΔH°f [Na2CO3(s)] = -1131 kJ

ΔH°f [CO2(g)] = -394 kJ

ΔH°f [H2O(l)] = -286 kJ

Answer 1

`DeltaH_f^@` `NaHCO_3(s)-=-840.9*kJ*mol^-1`

Explanation:

We want the enthalpy associated with the following reaction...

`Na(s) + C(s) + 1/2H_2(g) + 3/2O_2(g) rarr NaHCO_3(s)`

And `DeltaH_"rxn"^@-=DeltaH_f^@"sodium bicarbonate."`

And reasonably clearly, we cannot approach this directly. However, we can work with the given equation...

`2NaHCO_3(s) rarr Na_2CO_3(s) + CO_2(g) + H_2O(l)`
`DeltaH_"rxn"^@-=-129.3*kJ*mol^-1`

Now `DeltaH_"rxn"^@-=DeltaH_f^@"products"-DeltaH_f^@"reactants"`

`-129.3*kJ*mol^-1={(-1131-394-286)-2xx(DeltaH_f^@"sodium bicarbonate.")}*kJ*mol^-1`

And so....

`1681.7*kJ*mol^-1=-2(DeltaH_f^@"sodium bicarbonate")`

And thus `DeltaH_f^@"sodium bicarbonate"=-(1681.7)/2*kJ*mol^-1`

`=-840.9*kJ*mol^-1`...

Anyway, please check my arifmetik. All care taken but no responsibility admitted....