Answer:
![-112.876J](https://tex.z-dn.net/?f=-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)](https://tex.z-dn.net/?f=16KNO_3%28s%29%20%2B%2024C%28s%29%20%2B%20S_8%28s%29%20%20%20%20%5Cto%2024CO_2%28g%29%20%2B%208N_2%28g%29%20%2B%208K_2S%28s%29)
Let us define
work as;
![w_{pv} = - P_{external} \triangle Volume](https://tex.z-dn.net/?f=w_%7Bpv%7D%20%3D%20-%20P_%7Bexternal%7D%20%20%5Ctriangle%20Volume)
where ![\triangle (Volume) = (V_{final} - V_{initial})](https://tex.z-dn.net/?f=%5Ctriangle%20%28Volume%29%20%3D%20%28V_%7Bfinal%7D%20-%20V_%7Binitial%7D%29)
External pressure is given as ![1.00](https://tex.z-dn.net/?f=1.00)
, 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.](https://tex.z-dn.net/?f=V_i%20%3D%200.)
To find the volume of the products, we need to first find the amount of moles of the product made from
using the molar mass of
which is 101.1032 g/mol
![2.40_gKNO_3 . {\frac{1molKNO_3}{101.1032_g}} = 0.0237molKNO_3](https://tex.z-dn.net/?f=2.40_gKNO_3%20.%20%7B%5Cfrac%7B1molKNO_3%7D%7B101.1032_g%7D%7D%20%3D%200.0237molKNO_3)
Now let us convert moles of
into moles of
and
using the stoichiometric ratios from our balanced equation of the reaction.
![0.0237molKNO_3 . {\frac{24molCO_2}{16molKNO_3}} = 0.0356molCO_2](https://tex.z-dn.net/?f=0.0237molKNO_3%20.%20%7B%5Cfrac%7B24molCO_2%7D%7B16molKNO_3%7D%7D%20%3D%200.0356molCO_2)
![0.0237molKNO_3 . {\frac{8molN_2}{16molKNO_3}} = 0.01185molN_2](https://tex.z-dn.net/?f=0.0237molKNO_3%20.%20%7B%5Cfrac%7B8molN_2%7D%7B16molKNO_3%7D%7D%20%3D%200.01185molN_2)
is not factored into the volume calculation because it is a solid.
Now let us also convert the moles of
and
into grams using their respective molar masses.
![0.0356molCO_2 . {\frac{44.01_g}{1molCO_2}} = 1.567_gCO_2](https://tex.z-dn.net/?f=0.0356molCO_2%20.%20%7B%5Cfrac%7B44.01_g%7D%7B1molCO_2%7D%7D%20%3D%201.567_gCO_2)
![0.01185molN_2 . {\frac{28.014_g}{1molN_2}} = 0.332_gN_2](https://tex.z-dn.net/?f=0.01185molN_2%20.%20%7B%5Cfrac%7B28.014_g%7D%7B1molN_2%7D%7D%20%3D%200.332_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](https://tex.z-dn.net/?f=1.567_gCO_2%20.%20%7B%5Cfrac%7B1L%7D%7B1.830_g%7D%7D%20%3D%200.856LCO_2)
![0.332_gN_2 . {\frac{1L}{1.165_g}} = 0.285LN_2](https://tex.z-dn.net/?f=0.332_gN_2%20.%20%7B%5Cfrac%7B1L%7D%7B1.165_g%7D%7D%20%3D%200.285LN_2)
Summing up the two volumes, we get the final volume
![0.856L + 0.258L = 1.114L = V_f](https://tex.z-dn.net/?f=0.856L%20%2B%200.258L%20%3D%201.114L%20%3D%20V_f)
Plugging everything into the
equation, we get:
![w_{pv} = -1atm(1.114L - 0L) = -1.114L.atm](https://tex.z-dn.net/?f=w_%7Bpv%7D%20%3D%20-1atm%281.114L%20-%200L%29%20%3D%20-1.114L.atm)
Finally, let us convert
into joules using the conversion rate of;
![1L.atm = 101.325J\\-1.114L.atm. {\frac{101.325J}{1L.atm}} = -112.876J](https://tex.z-dn.net/?f=1L.atm%20%3D%20101.325J%5C%5C-1.114L.atm.%20%7B%5Cfrac%7B101.325J%7D%7B1L.atm%7D%7D%20%3D%20-112.876J)