Question

hello brainlians
today I've two questions. answer only if you know. one is of chemistry, and 1 is of maths.

read out the below given paragraph

the standard reaction Gibbs energy, ∆ᵣG⁰ of this reaction is positive. at the start of this reaction, there is one mole of χ₂ and no χ, as reaction proceeds, the number of moles of χ formed is given by β.

now answer the question

1) the equilibrium constant K_p for this reacting at 298K, in terms of β_{equilibrium} is?

maths Question:-

go through the attachment

now which of the following statements is true? answer in option(s)

(A) f(½) ≥ f(1)
(B) f(⅓) ≤ f(⅔)
(C) f'(2) ≤ 0
(D) f'(3)/f(3) ≥ f'(2)/f(2)

​

Answer 1

Answer for Question (1)

$\sf \begin{array}{ccc} \sf &\sf X_2 (g) & \rightleftharpoons & \sf X(g) \\\sf Initial \ mole: &\sf1 \ & \: & \sf 0 \\\sf t = t_{eq.} & \sf(1-\boldsymbol \alpha) & \: & 2\boldsymbol\alpha\end{array}$

Given: $\sf 2\boldsymbol\alpha = \boldsymbol\beta_{equilibrium}$

$\sf \implies \alpha = \dfrac{\beta_{equilibrium}}{2}$

total mole at equilibrium $\sf = (1+\alpha) = \left( 1 + \dfrac{1+\beta_{equilibrium}}{2} \right)$

$\sf \implies P_{x_{2}} = \left[ \dfrac{1- \frac{\beta_{equilibrium}}{2}}{1+\frac{\beta_{equilibrium}}{2}} \ P_{total}\right]$

$\sf = \left[ \dfrac{2-\beta_{equilibrium}}{2+\beta_{equilibrium}} \ P_{total}\right]$

$\sf P_{x(g)} = \left[ \dfrac{\beta_{equilibrium}}{1+ \dfrac{\beta_{equilibrium}}{2}} \right]$

$\sf = \dfrac{2 \beta_{equilibrium}}{2+\beta_{equilibrium}} P_{total}$

$\sf \therefore K_p = \dfrac{(Px)^2}{(Px_2)}$

$\sf = \dfrac{\left[ \dfrac{2- \beta_{equilibrium}}{2+\beta_{equilibrium}} \times P_{total}\right]^2}{\left[ \dfrac{2- \beta_{equilibrium}}{2+\beta_{equilibrium}} \times P_{total}\right]}$

$\sf = \dfrac{4\beta^2_{equilibrium}}{4-\beta^2_{equilibrium}} \times P_{total}$

$\boxed{\sf = \dfrac{8\beta^2_{equilibrium}}{4-\beta^2_{equilibrium}}}$

Answer for question (2)

$\displaystyle\sf f(x) = \lim_{n\to \infty} \left( \dfrac{\prod\limits_{r=1}^{n} \left(1+\dfrac{rx}{n}\right)}{\prod\limits_{r=1}^{n} \left( 1 + \left(\dfrac{rx}{n}\right)^2 \right) } \right)^{\dfrac{x}{n}}$

$\displaystyle\sf = e^{\left(\displaystyle\sf\int\limits_0^1 \ ln(1+xy)dy-ln(1+(xy)^2)dy\right)}$

$\displaystyle\sf = e^{\left(\displaystyle\sf\int\limits_0^x\ ln(1+t)dt - ln(1+t^2) dt\right)}$

$\displaystyle\sf\implies f'(x) = f(x) \ ln\left( \dfrac{1+x}{1+x^2}\right)$

for $\sf x\in (0,1)$ is increasing function

$\sf f'(2) = f(2) \ ln\left(\dfrac{3}{5}\right) < 0$

$\sf \implies \dfrac{f'(2)}{f(3)} = ln\left(\dfrac{2}{5} \right)$

$\sf \implies \dfrac{f'(2)}{f(2)} = ln\left(\dfrac{3}{5}\right)$

so options B,C are correct.

Answer 2

$Initial mole:t=teq.X2(g)1 (1−α)⇌X(g)02α$

$Given: \sf 2\boldsymbol\alpha = \boldsymbol\beta_{equilibrium}2α=βequilibrium</p><p></p><p>\sf \implies \alpha = \dfrac{\beta_{equilibrium}}{2}⟹α=2βequilibrium</p><p></p><p>$

total mole at equilibrium \sf = (1+\alpha) = \left( 1 + \dfrac{1+\beta_{equilibrium}}{2} \right)=(1+α)=(1+

2

1+β

equilibrium

)

\sf \implies P_{x_{2}} = \left[ \dfrac{1- \frac{\beta_{equilibrium}}{2}}{1+\frac{\beta_{equilibrium}}{2}} \ P_{total}\right]⟹P

x

2

=[

1+

2

β

equilibrium

1−

2

β

equilibrium

P

total

]

\sf = \left[ \dfrac{2-\beta_{equilibrium}}{2+\beta_{equilibrium}} \ P_{total}\right]=[

2+β

equilibrium

2−β

equilibrium

P

total

]

\sf P_{x(g)} = \left[ \dfrac{\beta_{equilibrium}}{1+ \dfrac{\beta_{equilibrium}}{2}} \right]P

x(g)

=

⎣

⎢

⎢

⎡

1+

2

β

equilibrium

β

equilibrium

⎦

⎥

⎥

⎤

\sf = \dfrac{2 \beta_{equilibrium}}{2+\beta_{equilibrium}} P_{total}=

2+β

equilibrium

2β

equilibrium

P

total

\sf \therefore K_p = \dfrac{(Px)^2}{(Px_2)}∴K

p

=

(Px

2

)

(Px)

2

\sf = \dfrac{\left[ \dfrac{2- \beta_{equilibrium}}{2+\beta_{equilibrium}} \times P_{total}\right]^2}{\left[ \dfrac{2- \beta_{equilibrium}}{2+\beta_{equilibrium}} \times P_{total}\right]}=

[

2+β

equilibrium

2−β

equilibrium

×P

total

]

[

2+β

equilibrium

2−β

equilibrium

×P

total

]

2

\sf = \dfrac{4\beta^2_{equilibrium}}{4-\beta^2_{equilibrium}} \times P_{total}=

4−β

equilibrium

2

4β

equilibrium

2

×P

total

\boxed{\sf = \dfrac{8\beta^2_{equilibrium}}{4-\beta^2_{equilibrium}}}

=

4−β

equilibrium

2

8β

equilibrium

2