Question

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Answer 1

$P_B: \texttt{\green{initial \: partial \: pressure \: of \: B}}$

$P_B': \texttt{\green{Final\: partial \: pressure \: of \: B}}$

$P_C: \texttt{\green{initial \: paryial \: pressure \: of \:C}}$

$P_C': \texttt{\green{Final \: paryial \: pressure \: of \: C}}$

Given

In the given reversible reaction,

$\mathsf{A(s) {\rightarrow} 2B(g) + 3C(g)}$

• $P_C' = 2P_C$

To Find:-

Final partial pressure of B ($P_B'$)

Explanation:-

$\mathsf{A(s) {\rightarrow} 2B(g) + 3C(g)}$

Note:-

□ Equilibrim constant(Kp or kc) remains constant.. even though there is a change in pressure

□ Eq. constant is only dependent on temp.

□ Assume the arrow.. as SPLITTED arrow

□ Active mass of Solids and liquids with gases is taken as unity

Now, W.k.t,

$\bold{K_p = [P_C]^{3} [P_B]^{2} }---(1)$

When, Partial pressure of C is doubled

$\bold{K_p = [P_C']^{3} [P_B']^{2}}---(2)$

Since, eq. constamt remains constant

(1) = (2)

${\rightarrow}[P_C]^{3} [P_B]^{2} =[P_C']^{3} [P_B']^{2}$

${\rightarrow}[P_C]^{3} [P_B]^{2} =[2P_C]^{3} [P_B']^{2}$

${\rightarrow}\cancel{[P_C]^{3}} [P_B]^{2} =8 \cancel{[P_C]^{3} }[P_B']^{2}$

${\rightarrow}[P_B]^{2} =8 [P_B']^{2}$

${\rightarrow}[P_B']^{2} = \dfrac{1}{8} [P_B']^{2}$

${\rightarrow}[P_B']= \dfrac{1}{2 \sqrt{2}}[P_B]$ _____option(3)