Answer:
A) The reaction is exothermic reaction
B) 46 gm CH3CH2OH = 1236KJ
15.3 gm ch3ch2OH = 1236/46 x 15.3
= 411.10 KJ..........released
therefore 1 KJ = 0.239 K cal
so, 411.10 x 0.239 kcal
= 98.2529 .................. released
c) 54 gm of H2O produced = 1236KJ
so, 42.7 gm H2O produced = 1236/54 x 42.7KJ
= 977.35 KJ released.
Explanation:
Answer:
17.6% is the percentage error
Explanation:
The percentage error is used to determine accuracy of a measurement (That is, how closely is the measure to the real or theoretical value). The equation is:
|Measure - Real| / Real * 100
In the problem, the measure was 22.7cm³ and real value is 19.3cm³. Solving for percentage error:
|22.7cm³ - 19.3cm³| / 19.3cm³ × 100
<h3>17.6% is the percentage error</h3>
Answer:
0.364
Explanation:
Let's do an equilibrium chart for the reaction of combustion of ammonia:
2NH₃(g) + (3/2)O₂(g) ⇄ N₂(g) + 3H₂O(g)
4.8atm 1.9atm 0 0 Initial
-2x -(3/2)x +x +3x Reacts (stoichiometry is 2:3/2:1:3)
4.8-2x 1.9-(3/2)x x 3x Equilibrium
At equilibrium the velocity of formation of the products is equal to the velocity of the formation of the reactants, thus the partial pressures remain constant.
If pN₂ = 0.63 atm, x = 0.63 atm, thus, at equilibrium
pNH₃ = 4.8 - 2*0.63 = 3.54 atm
pO₂ = 1.9 -(3/2)*0.63 = 0.955 atm
pH₂O = 3*0.63 = 1.89 atm
The pressure equilibrium constant (Kp) is calculated with the partial pressure of the gases substances:
Kp = [(pN₂)*(pH₂O)³]/[(pNH₃)²*
]
Kp = [0.63*(1.89)³]/[(3.54)²*
]
Kp = 4.2533/11.6953
Kp = 0.364