Question

Hello..


Question ; 1


Question;

Which of the following is incorrect?



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Answer 1

Hello mate !!

Thanks for asking this question !!

your answer is =>

let our given chemical reaction is =>

$m_{1}A_{1}+m_{2}A_{2}<---->n_{1}B_{1}+n_{2}B_{2}$

let the partial pressure of various reactants and products be =>

$pA_{1}, pA_{2}\: and\: pB_{1}, pB_{2}$

$K_p\: =\: \frac{(pB_{1})^{n_{1}}(pB_{2})^{n_{2}}}{(pA_{1})^{m_{1}}(pA_{2})^{m_{2}}}$

value of $K_{c}$ =>

$K_{c}\: =\: \frac{[B_{1}]^{n_{1}}[B_{2}]^{n_{2}}}{[A_{1}]^{m_{1}}[A_{2}]^{m_{2}}}$

For an ideal gas ,

PV = nRT

$P\: =\: active\: mass\:\times\: RT$

substitute value of partial pressure ,

$K_{p}\:=\:\frac{([B_{1}]^{n1}[B_{2}]^{n_{2}}RT^{(n_{1}+n_{2})})}{([A_{1}]^{m1}[A_{2}]^{m_{2}}RT^{(m_{1}+m_{2})})}$

Hence ,

$K_{p}\:=\:K_{c}\times{\frac{(RT)^{\sum{n}}}{(RT)^{\sum{m}}}}$

$K_{p}\:=\:K_{c}(RT)^{\sum{n}-\sum{m}}$

$\boxed{\bf{K_{p}\: =\: K_{c}(RT)^{\triangle{N}}}}$

by changing the sides of LHS and RHS you will get different formulae.

but, from the formula that we have proved here, proves that

option B is wrong.



$\underline{\huge\bf{\mathbb{B}e\mathbb{B}rainly}}$

Answer 2

Option 2 is incorrect ..............