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Nady [450]
3 years ago
13

Consider the formation of ammonia in two experiments. (a) to a 1.00-l container at 727°c, 1.30 mol of n2 and 1.65 mol of h2 are

added. at equilibrium, 0.100 mol of nh3 is present. calculate the equilibrium concentrations of n2 and h2 for the reaction: 2 nh3(g) n2(g) + 3 h2(g). n2 1.25 mol/l h2 1.50 mol/l find kc for the reaction. kc = 422 m2 (b) in a different 1.00-l container at the same temperature, equilibrium is established with 8.34 ✕ 10−2 mol of nh3, 1.50 mol of n2, and 1.25 mol of h2 present. calculate kc for the reaction: nh3(g) 1 2 n2(g) + 3 2 h2(g). 20.5 m (c) what is the relationship between the kc values in parts (a) and (b)? why aren't these values the same?
Chemistry
1 answer:
emmasim [6.3K]3 years ago
8 0
When it comes to equilibrium reactions, it useful to do ICE analysis. ICE stands for Initial-Change-Equilibrium. You subtract the initial and change to determine the equilibrium amounts which is the basis for Kc. Kc is the equilibrium constant of concentration which is just the ratio of products to reactant. 

Let's do the ICE analysis

      2 NH₃ ⇄ N₂ + 3 H₂
I         0        1.3    1.65
C     +2x       -x      -3x
-------------------------------------
E       0.1        ?        ?

The variable x is the amount of moles of the substances that reacted. You apply the stoichiometric coefficients by multiplying it by x. Now, we can solve x by:

Equilibrium NH₃ = 0.1 = 0 + 2x
x = 0.05 mol
Therefore,
Equilibrium H₂ = 1.65 - 3(0.05) = 1.5 mol
Equilibrium N₂ = 1..3 - 0.05 = 1.25 mol

For the second part, I am confused with the given reaction because the stoichiometric coefficients do not balance which violates the law of conservation of mass. But you should remember that the Kc values might differ because of the stoichiometric coefficient. For a reaction: aA + bB ⇄ cC, the Kc for this is

K_{C} = \frac{[ C^{c} ]}{[ A^{a} ][ B^{b} ]}

Hence, Kc could vary depending on the stoichiometric coefficients of the reaction.
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What is Keq for the reaction N₂ + 3H2 = 2NH3 if the equilibrium
hram777 [196]

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 \rm \dfrac{[NH_3]^2}{[N_2]\; [H_2]^3}

The given concentrations of the compounds have been:

Ammonia = 3 M

Nitrogen = 1 M

Hydrogen = 2 M

\rm \dfrac{9}{1\times 8} = 1.5

Thus, the correct option is D. Keq = 1.5.

Learn more about Keq

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3 0
2 years ago
Which nitrogen-containing base is found only in rna?
bija089 [108]
The answer to this is Codon. 
3 0
3 years ago
How could you verify that you produced carbon dioxide in your combustion reaction? 2. what indication did you have that nh3 was
DanielleElmas [232]

1) Carbon dioxide is a gas, so when CO_{2} is evolved in the reaction, it appears as bubbles. The gas released extinguishes the fire and it can turn lime water milky.

Ca(OH)_{2} (aq) + CO_{2}(g) ----> CaCO_{3}(s) +H_{2}O(l)

2) When NH_{3} is released in a decomposition reaction we can identify by the strong pungent smell of the gas released.

3) Saturated citric acid can cause corrosion of the metal layers present in the pipes. So, before draining out any acid it is neutralized so that the pipes and other plumbing works do not get damaged leading to leaks in the drainage system.

5 0
3 years ago
Which example of variation within species (select all correct answers)
s344n2d4d5 [400]
3 and 4 because the other two answers are different species.
6 0
3 years ago
Read 2 more answers
On a cool morning (12"C), a balloon is filled with 1.5 L of helium. By mid afternoon, the temperature has soared to 32°C. What i
ddd [48]

Answer:

1.6 L

Explanation:

Using Charle's law  

\frac {V_1}{T_1}=\frac {V_2}{T_2}

Given ,  

V₁ = 1.5 L

V₂ = ?

T₁ = 12 °C

T₂ = 32 °C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (12 + 273.15) K = 285.15 K  

T₂ = (32 + 273.15) K = 305.15 K  

Using above equation as:

\frac{1.5}{285.15}=\frac{V_2}{305.15}

V_2=\frac{1.5\cdot \:305.15}{285.15}

New volume = 1.6 L

8 0
3 years ago
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