The heat of reaction : 50.6 kJ
<h3>Further explanation</h3>
Based on the principle of Hess's Law, the change in enthalpy of a reaction will be the same even though it is through several stages or ways
Reaction
N₂(g) + 2H₂(g) ⇒N₂H₄(l)
thermochemical data:
1. N₂H₄(l)+O₂(g)⇒N₂(g)+2H₂O(l) ΔH=-622.2 kJ
2. H₂(g)+1/2O₂(g)⇒H₂O(l) ΔH=-285.8 kJ
We arrange the position of the elements / compounds so that they correspond to the main reaction, and the enthalpy sign will also change
1. N₂(g)+H₂O(l) ⇒ N₂H₄(l)+O₂(g) ΔH=+622.2 kJ
2. H₂(g)+1/2O₂(g)⇒H₂O(l) ΔH=-285.8 kJ x 2 ⇒
2H₂(g)+O₂(g)⇒2H₂O(l) ΔH=-571.6 kJ
Add reaction 1 and reaction 2, and remove the same compound from different sides
1. N₂(g)+2H₂O(l) ⇒ N₂H₄(l)+O₂(g) ΔH=+622.2 kJ
2.2H₂(g)+O₂(g)⇒2H₂O(l) ΔH=-571.6 kJ
-------------------------------------------------------------------- +
N₂(g) + 2H₂(g) ⇒N₂H₄(l) ΔH=50.6 kJ
Answer:
International System of Units
Definition: A system of physical units ( SI units ) based on the metre, kilogram, second, ampere, kelvin, candela, and mole, together with a set of prefixes to indicate multiplication or division by a power of ten.
Hope this helps!
Answer:
The number of moles of ammonia produced are 0.2 moles.
Explanation:
Given data:
Moles of N₂ = 0.100 mol
Moles of H₂ = 0.330 mol
Moles of NH₃ = ?
Solution:
First of all we will write the balance chemical equation
N₂ + 3H₂ → 2NH₃
Now we compare the moles of ammonia with nitrogen and hydrogen gas from balance chemical equation.
N₂ : NH₃
1 : 2
0.100 : 2/1 × 0.100 = 0.2 mol
H₂ : NH₃
3 : 2
0.330 : 2/3× 0.330 = 0.22 mol
The number of moles of ammonia produced by nitrogen gas are less so nitrogen will be limiting reactant and number of moles of ammonia produced are 0.2 moles.
Thw answer is PHj78 JJ CP30 R2D2
Answer:
The heat of combustion for the unknown hydrocarbon is -29.87 kJ/mol
Explanation:
Heat capacity of the bomb calorimeter = C = 1.229 kJ/°C
Change in temperature of the bomb calorimeter = ΔT = 2.19°C
Heat absorbed by bomb calorimeter = Q
Moles of hydrocarbon burned in calorimeter = 0.0901 mol
Heat released on combustion = Q' = -Q = -2,692 kJ
The heat of combustion for the unknown hydrocarbon :