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Irina18 [472]
3 years ago
7

The equilibrium constant for the reaction

Chemistry
2 answers:
avanturin [10]3 years ago
7 0

Equilibrium concentration of SO3=1.7M

Equilibrium concentration of SO2=0.070 M

Equilibrium concentration of NO2=1.3 M

Equilibrium constant is given as=10.8

The equilibrium reaction is given as:

SO2(g)+NI2(g)⇌NO(g) +SO3(g)

Keq= ([NO][SO3])/([SO2][NO2])

To calculate concentration of NO rearranging the above equation will give:

NO=Keq*[SO2][NO2]/[SO3]

NO=(10.8*0.070*1.3)/1.7

= 0.578M

So concentration of NO will be 0.578M

Veseljchak [2.6K]3 years ago
5 0

Answer:

The equilibrium concentration of NO in the gas mixture should be 0.578 M

Explanation:

We have equilibrium data for the chemical reaction in the gas phase:

SO2(g)+NO2(g)⇌NO(g) +SO3(g)

The equilibrium constant (Keq) for this reaction is 10.8.

Data for the components involved in this equilibrium are:

[SO3] = 1.7 M

[NO] = Unknown

[NO2] = 1.3 M

[SO2] = 0.07 M

We can express the Equilibrium constant as the multiplication of the concentrations of products divided by the multiplication of the concentration of reagents. After that, we can isolate [NO] concentration and find its value.

Keq = \frac{[NO][SO3]}{[SO2][NO2]}\\Keq *[SO2][NO2] = [NO][SO3]

[NO] = Keq \frac{[SO2][NO2]}{[SO3]}

Replacing the values given in the problem

[NO] = (10.8) \frac{[0.07M][1.3M]}{[1.7M]}  = 0.578 M

Finally, the equlibrium concentration is [NO] = 0.578 M

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Answer:

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Explanation:

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4 0
3 years ago
At constant pressure, which of these systems do work on the surroundings? A ( s ) + B ( s ) ⟶ C ( g ) A(s)+B(s)⟶C(g) 2 A ( g ) +
Tju [1.3M]

Correct question:

At constant pressure, which of these systems do work on the surroundings?

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

(c) A ( g ) + B ( g ) ⟶ C ( g )

(d) 2 A ( g ) + 2 B ( g ) ⟶ 3 C ( g )

Answer:

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

Explanation:

Work done by a system on the surroundings at a constant pressure is given as;

W = -PΔV

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ΔV is gas expansion, that is final volume of the gas minus initial volume of the gas must be greater than zero.

Part (a)

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Part (b)

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ΔV = 5 - ( 2+ 2) = 1 (expansion)

Part (c)

A ( g ) + B ( g ) ⟶ C ( g )

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Part (d)

2 A ( g ) + 2 B ( g ) ⟶ 3 C ( g )

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Thus, systems where there is gas expansion are in part (a) and part (b). The correct answers are:

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4 0
3 years ago
It took 55 days for a radioactivity of 1.75 x 1012 Bq to remain 0.135 Ci. What is the half-life of this radioactivity?
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From the calculations, the half life of the material is 6.5 days.

<h3>What is radioactivity?</h3>

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Uisng the formula;

N=Noe^-kt

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No = amount initially present =  1.75 x 10^12 Bq

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t = time taken = 55 days

Hence;

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Learn more about radioactivity:brainly.com/question/1770619

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