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uysha [10]
4 years ago
7

A mixture of 0.10 molmol of NONO, 0.050 molmol of H2H2, and 0.10 molmol of H2OH2O is placed in a 1.0-LL vessel at 300 KK. The fo

llowing equilibrium is established:
2NO(g)+2H2(g)←→N2(g)+2H2O(g)2NO(g)+2H2(g)←→N2(g)+2H2O(g)
At equilibrium [NO]=0.062M[NO]=0.062M.

Part A:

Calculate the equilibrium concentration of H2

Part B:

Calculate the equilibrium concentration of N2N2.

Part C:

Calculate the equilibrium concentration of H2OH2O.

Part D:

Calculate KcKc.
Chemistry
1 answer:
vitfil [10]4 years ago
8 0

Answer:

A) [H2] = 0.012 M

B) [N2]  = 0.019M

C) [H2O] = 0.138  M

D) Kc = 653.6

Explanation:

Step 1: Data given

Number of moles NO = 0.10 moles

Number of moles H2 = 0.050 moles

Number of moles H2O = 0.10 moles

Volume = 1.0 L

Temperature = 300 K

At equilibrium [NO]=0.062M

Step 2: The balanced equation

2NO(g)+2H2(g)←→N2(g)+2H2O(g)

Step 3: The initial  concentrations

Concentrations = moles / volume

[NO] = 0.10 moles / 1.0 L = 0.10 M

[H2] = 0.050 moles / 1.0 L = 0.050 M

[H2O] = 0.10 moles / 1.0 L = 0.10 M

[N2] = 0 M

Step 4: The concentration at the equilibrium

For 2 moles NO we need 2 moles H2 to produce 1 mol N2 and 2 moles H2O

[NO] = 0.10 - 2X

[H2] = 0.050 - 2X M

[H2O] = 0.10 + 2X M

[N2] = X

[NO] = 0.10 - 2X M = 0.062

X = 0.019

[NO] = 0.062M

[H2] = 0.050 - 2*0.019 = 0.012 M

[H2O] = 0.10 + 2*0.019 = 0.138  M

[N2] = X = 0.019M

Step 5: Calculate Kc

Kc = [N2][H2O]² / [NO]²[H2]²

Kc = (0.019*0.138²)/(0.062²*0.012²]

Kc = 653.6

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2 years ago
Determine whether the following reaction is a Lewis acid/base reaction and if so, what is the Lewis acid: Fe3+(aq) + 6 CN−(aq) →
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Answer: It is a Lewis acid/base reaction, and Fe^{3+} is the Lewis acid.

Explanation:

According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.

Fe^{3+} can readily accept electrons and thus act as a lewis acid which is short of electrons.

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4 years ago
What is the rate law for the reaction 2A + 2B + 2C --&gt; products
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Answer:

R = 47.19 [A]*([B]^2)*[C]

Explanation:

The rate law for the reaction 2A + 2B + 2C --> products

Is being sought.

The reaction rate R could be expressed as  

R = k ([A]^m)*([B]^n)*([C]^p)                      (1)

where m, n, and p are the reaction orders with respect to (w.r.t.) components A, B and C respectively. This could be reduced to

R = ka ([A]^m)                   (2)

Where ka=(k[B]^n)*([C]^p);    

R = kb ([B]^n)                    (3)

Where kb=(k[A]^m)*([C]^p); and  

R = kc ([C]^p)                     (4)

Where kc=(k[A]^m)*([B]^n).

Equations (2), (3) and (4) are obtained for cases when the concentrations of two components are kept constant, while only one component’s concentration is varied. We can determine the reaction wrt each component by employing these equations.  

The readability is very much enhanced when the given data is presented in the following manner:

Initial [A]  0.273   0.819   0.273   0.273

Initial [B]  0.763   0.763   1.526   0.763

Initial [C]  0.400   0.400    0.400   0.800

Rate           3.0       9.0       12.0       6.0

Run#  1  2 3  4

Additional row is added to indicate the run # for each experiment for easy reference.

First, we use the initial rate method to evaluate the reaction order w.r.t. each component [A], [B] and [C] based on the equations (2), (3) and (4) above.

Let us start with the order wrt [A]. From the given data, for experimental runs 1 and 2, the concentrations of reactants B and C were kept constant.

Increasing [A] from 0.273 to 0.819 lead to the change of R from 3.0 to 9.0, hence we can apply the relation based on equation (2) between the final rate R2, the initial rate R1 and the final concentration [A2] and the initial concentration [A1] as follows:

R2/R1=ka[A2]^m/ka[A1]^m=([A2]/[A1])^m

9.0/3.0 = (0.819/0.273)^m

3 = (3)^m = 3^1  -> m = 1

Similarly, applying experimental runs 1 and 3 could be applied for the determination of n, by employing equation (3):  

R3/R1=kb[B3]^n/kb[B1]^n=([B3]/[B1])^n

12/3= (1.526/0.763)^n

4= 2^n, -> n = 2

And finally for the determination of p we have using runs 4 and 1:

R4/R1=kc[C4]^p/kc[C1]^p=([C4]/[C1])^p

6/3= (0.8/0.4)^p

2= 2^p , -> p = 1

Therefore, plugging in the values of m, n and p into equation (1), the rate law for the reaction will be:

R = k [A]*([B]^2)*[C]

The value of the rate constant k could be estimated by making it the subject of the formula, and inserting the given values, say in run 1:

k = R /( [A]*([B]^2)*[C]) = 3/0.273*(0.763^2)*0.4 =

47.19

Finally, the rate law is

R = 47.19 [A]*([B]^2)*[C]

7 0
4 years ago
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Answer:

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Explanation:

Given parameters:

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Unknown:

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