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lyudmila [28]
3 years ago
6

Numerator and denominator are Equal to one another? I don’t get the concept

Chemistry
1 answer:
Alekssandra [29.7K]3 years ago
5 0

It's like saying that if 1 meter is equal to 100 cm,

\frac{100cm}{1m}

or

\frac{1m}{100cm}

You can base the answers from the chart

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Consider the following reaction: A(g)⇌2B(g). Find the equilibrium partial pressures of A and B for each of the following differe
Illusion [34]

Answer:

a. Kp=1.4

P_{A}=0.2215 atm

P_{B}=0.556 atm

b.Kp=2.0 * 10^-4

P_{A}=0.495atm

P_{B}=0.00995 atm

c.Kp=2.0 * 10^5

P_{A}=5*10^{-6}atm

P_{B}=0.9999 atm

Explanation:

For the reaction  

A(g)⇌2B(g)

Kp is defined as:

Kp=\frac{(P_{B})^{2}}{P_{A}}

The conditions in the system are:

          A                    B

initial   0                1 atm

equilibrium x       1atm-2x

At the beginning, we don’t have any A in the system, so B starts to react to produce A until the system reaches the equilibrium producing x amount of A. From the stoichiometric relationship in the reaction we get that to produce x amount of A we need to 2x amount of B so in the equilibrium we will have 1 atm – 2x of B, as it is showed in the table.    

Replacing these values in the expression for Kp we get:

Kp=\frac{(1-2x)^{2}}{x}

Working with this equation:

x*Kp=(1-2x)^{2} - -> x*Kp=4x^{2}-4x+1- - >4x^{2}-(4+Kp)*x+1=0

This last expression is quadratic expression with a=4, b=-(4+Kp) and c=1

The general expression to solve these kinds of equations is:

x=\frac{-b(+-)*\sqrt{(b^{2}-4ac)}}{2a} (equation 1)

We just take the positive values from the solution since negative partial pressures don´t make physical sense.

Kp = 1.4

x_{1}=\frac{(1.4+4)+\sqrt{(-(1.4+4)^{2}-4*4*1)}}{2*4}=1.128

x_{1}=\frac{(1.4+4)-\sqrt{(-(1.4+4)^{2}-4*4*1)}}{2*4}=0.2215

With x1 we get a partial pressure of:

P_{A}=1.128 atm

P_{B}=1-2*1.128 = -1.256 atm

Since negative partial pressure don´t make physical sense x1 is not the solution for the system.

With x2 we get:

P_{A}=0.2215 atm

P_{B}=1-2*0.2215 = 0.556 atm

These partial pressures make sense so x2 is the solution for the equation.

We follow the same analysis for the other values of Kp.

Kp=2*10^-4

X1=0.505

X2=0.495

With x1

P_{A}=0.505atm

P_{B}=1-2*0.505 = -0.01005 atm

Not sense.

With x2

P_{A}=0.495atm

P_{B}=1-2*0.495 = 0.00995 atm

X2 is the solution for this equation.  

Kp=2*10^5

X1=50001

X2=5*10^{-6}

With x1

P_{A}=50001atm

P_{B}=1-2*50001=-100001atm

Not sense.

With x2

P_{A}= 5*10^{-6}atm

P_{B}=1-2*5*10^{-6}= 0.9999 atm

X2 is the solution for this equation.  

8 0
3 years ago
Which is a property of an ideal gas?
AlekseyPX
HEY THERE!

THE ANSWER IS: the properties of an ideal gas are: An ideal gas consists of a large number of identical molecules. The volume occupied by the molecules themselves is negligible compared to the volume occupied by the gas. The molecules obey Newton's laws of motion, and they move in random motion.

CREDITS:
<span>physics.bu.edu/~duffy/py105/Idealgas.htm</span>
6 0
3 years ago
Read 2 more answers
What chemical reaction involve acid and bases​
svetoff [14.1K]

Answer:

Neutralization reaction

5 0
3 years ago
Read 2 more answers
What are alternative periodic tables
aksik [14]
Tabulations of chemical elements differing in their organization from the traditional seen periodic system
5 0
3 years ago
If a substance has a half life of 58 years and starts with 500 g radioactive, how much remains radioactive after 30 years?
Vilka [71]

Answer:

A = 349 g.

Explanation:

Hello there!

In this case, since the radioactive decay kinetic model is based on the first-order kinetics whose integrated rate law is:

A=Ao*exp(-kt)

We can firstly calculate the rate constant given the half-life as shown below:

k=\frac{ln(2)}{t_{1/2}} =\frac{ln(2)}{58year}=0.012year^{-1}

Therefore, we can next plug in the rate constant, elapsed time and initial mass of the radioactive to obtain:

A=500g*exp(-0.012year^{-1} *30year)\\\\A=349g

Regards!

5 0
3 years ago
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