Question

At a certain temperature total pressure of 1atm ,a2 vapours contains 20 %by volume of a atom a2 gives 2 a find kp for this reaction

Answer 1

Answer: The $K_p$ of the reaction is 0.05

Explanation:

We are given:

20 % volume of A atoms

So, percentage by volume of $A_2$ molecules = 100 - 20 = 80 %

We know that: Moles is equivalent to volume of the gas

$V\propto n$    (Avogadro's law)

Now, the mole fraction of A atoms = $\frac{20}{100}=0.2$

The mole fraction of $A_2$ molecules = $\frac{80}{100}=0.8$

The partial pressure of a gas is given by Raoult's law, which is:

$p_A=p_T\times \chi_A$     ......(1)

where,

$p_A$ = partial pressure of substance A

$p_T$ = total pressure

$\chi_A$ = mole fraction of substance A

• For A atoms:

We are given:

$p_t=1atm\\\chi_A=0.2$

Putting values in equation 1, we get:

$p_A=0.2\times 1=0.2atm$

• For $A_2$ molecules:

We are given:

$p_t=1atm\\\chi_{A_2}=0.8$

Putting values in equation 1, we get:

$p_{A_2}=0.8\times 1=0.8atm$

For the given chemical reaction:

$A_2\rightarrow 2A$

The expression of $K_p$ for the above reaction follows:

$K_p=\frac{p_{A}^2}{p_{A_2}}$

Putting values in above equation, we get:

$K_p=\frac{(0.2)^2}{0.8}\\\\K_p=0.05$

Hence, the $K_p$ of the reaction is 0.05

Answer 2

Explanation:

Answer: The K_pK

p

of the reaction is 0.05

Explanation:

We are given:

20 % volume of A atoms

So, percentage by volume of A_2A

2

molecules = 100 - 20 = 80 %

We know that: Moles is equivalent to volume of the gas

V\propto nV∝n (Avogadro's law)

Now, the mole fraction of A atoms = \frac{20}{100}=0.2

100

20

=0.2

The mole fraction of A_2A

2

molecules = \frac{80}{100}=0.8

100

80

=0.8

The partial pressure of a gas is given by Raoult's law, which is:

p_A=p_T\times \chi_Ap

A

=p

T

×χ

A

......(1)

where,

p_Ap

A

= partial pressure of substance A

p_Tp

T

= total pressure

\chi_Aχ

A

= mole fraction of substance A

For A atoms:

We are given:

\begin{lgathered}p_t=1atm\\\chi_A=0.2\end{lgathered}

p

t

=1atm

χ

A

=0.2

Putting values in equation 1, we get:

p_A=0.2\times 1=0.2atmp

A

=0.2×1=0.2atm

For A_2A

2

molecules:

We are given:

\begin{lgathered}p_t=1atm\\\chi_{A_2}=0.8\end{lgathered}

p

t

=1atm

χ

A

2

=0.8

Putting values in equation 1, we get:

p_{A_2}=0.8\times 1=0.8atmp

A

2

=0.8×1=0.8atm

For the given chemical reaction:

A_2\rightarrow 2AA

2

→2A

The expression of K_pK

p

for the above reaction follows:

K_p=\frac{p_{A}^2}{p_{A_2}}K

p

=

p

A

2

p

A

2

Putting values in above equation, we get:

\begin{lgathered}K_p=\frac{(0.2)^2}{0.8}\\\\K_p=0.05\end{lgathered}

K

p

=

0.8

(0.2)

2

K

p

=0.05

Hence, the K_pK

p

of the reaction is 0.05