Answer:
[N2] = 0.3633M
[H2] = 1.090M
[NH3] = 0.2734M
Explanation:
Based on the reaction of the problem, Kc is defined as:
Kc = 0.159 = [NH3]² / [N2] [H2]³
<em>Where [] are the equilibrium concentrations.</em>
The initial concentrations of the reactants is:
N2 = 1.00mol / 2.00L = 0.500M
H2 = 3.00mol / 2.00L = 1.50M
When the equilibrium is reached, the concentrations are:
[N2] = 0.500M - X
[H2] = 1.50M - 3X
[NH3] = 2X
<em>Where X is reaction quotient</em>
Replacing in the Kc equation:
0.159 = [2X]² / [0.500 - X] [1.50 - 3X]³
0.159 = 4X² / 1.6875 - 13.5 X + 40.5 X² - 54 X³ + 27 X⁴
0.268313 - 2.1465 X + 6.4395 X² - 8.586 X³ + 4.293 X⁴ = 4X²
0.268313 - 2.1465 X + 2.4395 X² - 8.586 X³ + 4.293 X⁴ = 0
Solving for X:
X = 0.1367. Right solution.
X = 1.8286. False solution. Produce negative concentrations
Replacing:
[N2] = 0.500M - 0.1367M
[H2] = 1.50M - 3*0.1367M
[NH3] = 2*0.1367M
The equilibrium concentrations are:
<h3>[N2] = 0.3633M</h3><h3>[H2] = 1.090M</h3><h3>[NH3] = 0.2734M</h3>
Answer:
ΔH = ΔH₁ + ΔH₂ - ΔH₃
Explanation:
Given that:
1. A → 2B
2. B → C + D
3. E → 2D
Assuming from the corresponding ΔH for process 1, 2 and 3 are ΔH₁, ΔH₂, ΔH₃ respectively.
To estimate the ΔH for the process A → 2C + E
We multiply 2 with equation 2 where (B → C + D)
2B → 2C + 2D ⇒ 2ΔH₂
Also, let's switch equation (3), such that we have,
2D → E -ΔH₃
The summation of all the equation result into :
A → 2C + E
where; ΔH = ΔH₁ + ΔH₂ - ΔH₃
It increases availability of food for humans
The balanced chemical reaction is:
<span>3N2H4(l)→4NH3(g)+N2(g)
</span>
The amounts given for the N2H4 reactant will be the starting point for our calculations.
2.6mol N2H4 ( 4 mol NH3 / 3 mol N2H4 ) = 3.47 mol NH3
4.05mol N2H4 ( 4 mol NH3 / 3 mol N2H4 ) = 5.4 mol NH3
63.8g N2H4 <span>( 4 mol NH3 / 3 mol N2H4 ) = 85.07 mol NH3</span>
Answer:
sdoawjdiowadawoi siokdwajsiokwjDIWIAdawidjaskmdnkawdjad kadakwdkawdawhdaw
Explanation: im smart
