Question

40.0 mL of 0.200 M aqueous NaOH is added to 200.0 mL of 0.100 M aqueous NaHCO3 in a flask maintained at 25 ?C. Neglecting the effects of dilution, what is q for this reaction?

Answer 1

What does q stand for?

Answer 2

Answer : The 'q' for this reaction is, 328 J

Explanation :

First we have to calculate the moles of $NaOH$ and $NaHCO_2$.

$\text{Moles of }NaOH=\text{Molarity of }NaOH\times \text{Volume of solution}=0.200mole/L\times 0.04L=0.008mole$

$\text{Moles of }NaHCO_2=\text{Molarity of }NaHCO_2\times \text{Volume of solution}=0.100mole/L\times 0.2L=0.02mole$

From this we conclude that, the moles of $NaOH$ are less than moles of $NaHCO_3$. So, the limiting reactant is, NaOH.

Now we have to calculate the 'q' for this reaction.

The balanced chemical reaction will be,

$OH^-+HCO_3^-\rightarrow CO_3^{2-}+H_2O$

The expression used for 'q' of this reaction is:

$q=n(\Delta H_f^o\text{ of product})-n(\Delta H_f^o\text{ of reactant})$

$q=n[(\Delta H_f^o\text{ of }CO_3^{2-})+(\Delta H_f^o\text{ of }H_2O)]-n[(\Delta H_f^o\text{ of }HCO_3^-)+(\Delta H_f^o\text{ of }OH^-)]$

where,

n = number of moles of limiting reactant = 0.008 mole

At room temperature,

$\Delta H_f^o\text{ of }CO_3^{2-}$ = -677 kJ/mole

$\Delta H_f^o\text{ of }H_2O$ = -286 kJ/mole

$\Delta H_f^o\text{ of }HCO_3^-$ = -692 kJ/mole

$\Delta H_f^o\text{ of }OH^-$ = -230 kJ/mole

Now put all the given values in the above expression, we get:

$q=0.008mole[(-677kJ/mole)+(-286kJ/mole)]-0.008mole[(-692kJ/mole)+(-230kJ/mole)]$

$q=0.328kJ=328J$      (conversion used : 1 kJ = 1000 J)

Therefore, the 'q' for this reaction is, 328 J