The Keq for the reaction N₂ + 3H2 = 2NH3 if the equilibrium concentrations are Keq = 1.5. The correct option is D.
<h3>What is Keq?</h3>
Keq is the ratio of the concentration of reactant to the concentration of the product.
The balanced equation is
N₂ + 3H₂ = 2NH₃
The equilibrium constant is
The given concentrations of the compounds have been:
Ammonia = 3 M
Nitrogen = 1 M
Hydrogen = 2 M
Thus, the correct option is D. Keq = 1.5.
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Answer: it honestly depends.
Explanation:
Answer:
D. Nuclei with small masses combine to form nuclei with larger masses.
B. A small amount of mass in the nuclei that combine is converted to energy
Explanation:
A nuclear fusion, in contrary to fission, is the process by which the nuclei of two atoms combine to form a much larger atom with a large nuclei. Likewise, during a fusion reaction, a large amount of energy is released from the small amount of mass in the nuclei (two) that combines.
According to this question, the following are true of a fusion reaction:
- Nuclei with small masses combine to form nuclei with larger masses.
- A small amount of mass in the nuclei that combine is converted to enormous energy.
Answer:
The conversion achieved for the first CSTR impeller is 0.382
Discrepancy = 0.188
Explanation:
The impeller divides the CSTR into 2 equal reactors of volume 500gal
Using V = FaoX/ (-ra)
500gal = Fao×Xa/[(KCao^2( 1 -X1)^2]
500gal = CaoVoX1/ KCao^2(1-X1)
500gal= 500gal × X1'/(1 - X1)^2
(1 -X1)^2 = X1
X1^2 - 3X1 + 1 = 0
X1= 0.382
Conversion achieved in the first CSTR is 0.382
Actual measured CSTR = 57% =57/100=0.57
Discrepancy in the conversions= 0.57 -0.383 =0.188
Explanation:
Relation between pressure, latent heat of fusion, and change in volume is as follows.
Also,
where, is the difference in specific volumes.
Hence,
As, = 22.0 J/mol K
And, ...... (1)
where, = density of water
= density of ice
M = molar mass of water =
Therefore, using formula in equation (1) we will calculate the volume of fusion as follows.
=
=
Therefore, calculate the required pressure as follows.
=
or, = 145 bar/K
Hence, for change of 1 degree pressure the decrease is 145 bar and for 4.7 degree change dP =
= 681.5 bar
Thus, we can conclude that pressure should be increased by 681.5 bar to cause 4.7 degree change in melting point.