<u>Answer:</u> The value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

<u>Explanation:</u>

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = 3.45

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-1=1$

Putting values in above equation, we get:

$3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084$

The equation used to calculate concentration of a solution is:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

Initial moles of $(CH_3)_3CCl(g)$ = 1.00 mol

Volume of the flask = 5.00 L

So, $\text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M$

For the given chemical reaction:

$(CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)$

Initial: 0.2 - -

At Eqllm: 0.2 - x x x

The expression of $K_c$ for above reaction follows:

$K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}$

Putting values in above equation, we get:

$0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178$

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

$[(CH_3)_2C=CH]=x=0.094M$

$[HCl]=x=0.094M$

$[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M$

Hence, the value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

8 0

3 years ago

One of the most difficult solar energy design problems to solve is:

Sonja [21]

I'm not sure but i would say C.storing the heat energy.

6 0

3 years ago

Read 2 more answers

Please answer all of the questions

anzhelika [568]

Here's your ans

Explanation:

1-1.50

2-d

3-c

4-b

5-d

3 0

2 years ago

You observe 2 waves named alpha and beta. The alpha wave has a frequency of 5 Hz and the beta wave has a frequency of 2 Hz. Base

MAVERICK [17]

Answer:

shorter wavelength = alpha wave

Explanation:

Given that,

The alpha wave has a frequency of 5 Hz and the beta wave has a frequency of 2 Hz.

We need to compare the wavelengths of these two waves.

For alpha wave,

$\lambda_1=\dfrac{c}{f_1}\\\\\lambda_1=\dfrac{3\times 10^8}{5}\\\\=6\times 10^7\ m$

For beta wave,

$\lambda_2=\dfrac{c}{f_2}\\\\\lambda_2=\dfrac{3\times 10^8}{2}\\\\=15\times 10^7\ m$

From the above calculations, we find that the wavelength of the alpha wave is shorter than the wavelength of the beta wave.

8 0

3 years ago