Question

If 2.40 g of KNO3 reacts with sufficient sulfur (S8) and carbon (C), how much P-V work will the gases do against an external pressure of 1.00 atm given the densities of nitrogen and carbon dioxide are 1.165 g/L and 1.830 g/L, respectively, at 20°C?

Answer 1

Answer:

$-112.876J$

Explanation:

In order to solve this question, we would need to incorporate Stoichiometry, which involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data.

Here's a balanced equation for the reaction:

$16KNO_3(s) + 24C(s) + S_8(s) \to 24CO_2(g) + 8N_2(g) + 8K_2S(s)$

Let us define $P - V$ work as;

$w_{pv} = - P_{external} \triangle Volume$

where  $\triangle (Volume) = (V_{final} - V_{initial})$

External pressure is given as  $1.00$$atm$ , therefore the work solely depends on the change in volume and since the reactants are solids, none of the reactants contribute to the volume. Hence,  $V_i = 0.$

To find the volume of the products, we need to first find the amount of moles of the product made from  $2.40_gKNO_3,$ using the molar mass of  $KNO_3$  which is 101.1032 g/mol  

$2.40_gKNO_3 . {\frac{1molKNO_3}{101.1032_g}} = 0.0237molKNO_3$

Now let us convert moles of  $KNO_3$  into moles of $CO_2$ and $N_2$  using the stoichiometric ratios from our balanced equation of the reaction.

$0.0237molKNO_3 . {\frac{24molCO_2}{16molKNO_3}} = 0.0356molCO_2$

$0.0237molKNO_3 . {\frac{8molN_2}{16molKNO_3}} = 0.01185molN_2$

$K_2S$ is not factored into the volume calculation because it is a solid.

Now let us also convert the moles of  $CO_2$  and  $N_2$ into grams using their respective molar masses.

$0.0356molCO_2 . {\frac{44.01_g}{1molCO_2}} = 1.567_gCO_2$

$0.01185molN_2 . {\frac{28.014_g}{1molN_2}} = 0.332_gN_2$

We will now proceed to convert grams into volume using the density values provided.

$1.567_gCO_2 . {\frac{1L}{1.830_g}} = 0.856LCO_2$

$0.332_gN_2 . {\frac{1L}{1.165_g}} = 0.285LN_2$

Summing up the two volumes, we get the final volume

$0.856L + 0.258L = 1.114L = V_f$

Plugging everything into the $w_{pv}$ equation, we get:

$w_{pv} = -1atm(1.114L - 0L) = -1.114L.atm$

Finally, let us convert $L.atm$ into joules using the conversion rate of;

$1L.atm = 101.325J\\-1.114L.atm. {\frac{101.325J}{1L.atm}} = -112.876J$