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Anastaziya [24]
3 years ago
13

A system was prepared with NH3 = O2 = 3.60 M as the only components initially. At equilibrium N2O4 is 0.60M. Calculate the value

of the equilibrium constant Kc for the reaction
Chemistry
1 answer:
frozen [14]3 years ago
8 0

Answer:

Kc = 0.0216

Explanation:

To do this, we need to write the equilibrium reaction which is the following:

NH3 + O2 <------> N2O4 + H2O

And now, let's balance the equation

4NH3 + 7O2 <--------> 2N2O4 + 6H2O

The general expression for Kc is:

Kc = [N2O4]²[H2O]⁶ / [O2]⁷[NH3]⁴

Now, in order to know the concentrations, we need to do an ICE chart here, so we can know the final concentrations of all species here:

         4NH3 + 7O2 <--------> 2N2O4 + 6H2O

I:          3.6        3.6                      0            0

C:         -4x       -7x                      +2x        +6x

E:      3.6 - 4x   3.6 - 7x              0.6          +6x

According to this, we know that N2O4 is 0.6M, but in the change we have that it gained 2x so the value of x will be:

2x = 0.6

x = 0.6/2 = 0.3

Therefore, the value of x is 0.3 ans we can know now the concentrations of all species in equilibrium, and then, the kc:

[NH3] = 3.6 - 4(0.3) = 2.4 M

[O2] = 3.6 - 7(0.3) = 1.5 M

[H2O] = 6(0.3) = 1.8 M

Now the value of Kc would be:

Kc = (0.6)² (1.8)⁶ / (2.4)⁴ (1.5)⁷

Kc = 0.0216

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g a solution is made by mixing 500.0 mL of 0.037980.03798 M Na2sO4 Na2sO4 with 500.0 mL of 0.034280.03428 M NaOH NaOH . Complete
Mandarinka [93]

Answer:

The concentration of the sodium and arsenate ions at the end of the reaction in the final solution

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Explanation:

Complete Question

A solution is made by 500.0 mL of 0.03798 M Na₂HAsO₄ with 500.0 mL of 0.03428 M NaOH. Complete the mass balance expressions for the sodium and arsenate species in the final solution.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

From the information provided in this question, we can calculate the number of moles of each reactant at the start of the reaction and we then determine which reagent is in excess and which one is the limiting reagent (in short supply and determines the amount of products to be formed)

Concentration in mol/L = (Number of moles) ÷ (Volume in L)

Number of moles = (Concentration in mol/L) × (Number of moles)

For Na₂HAsO₄

Concentration in mol/L = 0.03798 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03798 × 0.5 = 0.01899 mole

For NaOH

Concentration in mol/L = 0.03428 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03428 × 0.5 = 0.01714 mole

Since the NaOH is in short supply, it is evident that it is the limiting reagent and Na₂HAsO₄ is in excess.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

0.01899        0.01714        0           0 (At time t=0)

(0.01899 - 0.1714) | 0 → 0.01714    0.01714 (end)

0.00185  | 0 → 0.01714    0.01714 (end)  

Hence, at the end of the reaction, the following compounds have the following number of moles

Na₂HAsO₄ = 0.00185 mole

This means Na⁺ has (2×0.00185) = 0.0037 mole at the end of the reaction and (HAsO₄)²⁻ has 0.00185 mole at the end of the reaction

NaOH = 0 mole

Na₃AsO₄ = 0.01714 moles

This means Na⁺ has (3×0.01714) = 0.05142 mole at the end of the reaction and (AsO₄)³⁻ has 0.01714 mole at the end of the reaction

H₂O = 0.01714 moles

So, at the end of the reaction

Na⁺ has 0.0037 + 0.05142 = 0.05512 mole

(HAsO₄)²⁻ has 0.00185 mole

(AsO₄)³⁻ has 0.01714 mole.

And since the Total volume of the reaction setup is now 500 mL + 500 mL = 1000 mL = 1 L

Hence, the concentration of the sodium and arsenate ions at the end of the reaction is

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Hope this Helps!!!

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2H2(g]+O2(g]→2H2O(l]]. Notice that the reaction requires 2 moles of hydrogen gas and 1

Explanation:

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