Question

2KNO3(s) + $$$(s) + 3C(s) — K_S(s) +N2(g) + 3C02(3) 2nd attempt Part 1 . See Periodic Table D See Hint If 2.70 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? Part 2 ♡ See Hint If the reaction produces 58.3 kJ of heat, what is A Efor the system?

Answer 1

Ans :-

From the balance chemical equation :

2 KNO₃ (s) + 1/8 S₈ (s) + 3 C (s) --------------> K₂S (s) + N₂ (g) + 3 CO₂ (g)

Given mass of KNO₃ = 2.70 g

Molar mass of KNO₃ = 101.1 g/mol

So,

Number of moles of KNO₃ = Given mass of KNO₃/Gram molar mass of KNO₃

= 2.70 g/101.1 g/mol

= 0.0267 mol

From the balance chemical equation :

Number of moles of CO₂ (g) = 3/2 x (Number of moles of KNO₃)

= 3/2 x 0.0267 mol mol

= 0.040 mol

Also,

Mass of CO₂ (g) = Moles x Gram molar mass of CO₂ (g)

= 0.040 mol x 44 g/mol

= 1.76 g

Volume of CO₂ (g) = Mass/Density = 1.76 g / 1.830 g/L = 0.962 L

Similarly,

Number of moles of N₂ (g) = 3/2 x (Number of moles of N₂)

= 1/2 x 0.0267 mol mol

= 0.01335 mol

Also,

Mass of N₂ (g) = Moles x Gram molar mass of N₂(g)

= 0.01335 mol x 28 g/mol

= 0.3738 g

Volume of N₂ (g) = Mass/Density = 0.3738 g / 1.165 g/L = 0.321 L

Because, K₂S is a solid, Therefore,

Change in Volume of gases becomes = ΔV = 0.962 L + 0.321 L= 1.283 L

We know,

w = -P_ext.ΔV

= - (1.00 atm).(1.283 L)

= -1.283 L atm

= -1.283 x 101.3 J/mol

= -130.0 J/mol

= - 0.130 KJ/mol

Hence, PV-work = - 0.130 KJ

Here, -ve sign indicates that work done by the gases.

Also,

Change in internal energy = ΔE = q + w

ΔE = q + w

= - 58.3 KJ - 0.130 KJ

= - 58.43 KJ

Hence, ΔE = - 58.43 KJ

Here, -ve sign indicates that heat evolved in this reaction.