Two solids dissociate as follows A (s ) B (g )+C (g ) ;K_P_1=x atm^2 D (s ) C (g )+E (g ) ;K_P_2=y atm^2 The total pressure when both the solids dissociate simultaneously is:

Question

Accepted Answer

Correct answer: B. (b) A (s ) ⟶B (g )P_1+C (g )P_1+P_2 K_P_1=P_B P_C P_1(P_1+P_2)=x . . . . . (i ) D (s ) ⟶C (g )P_1+P_2+E (g )P_2 K_P_2=P_C P_E (P_1+P_2)P_2=y . . . . . ( ii ) Adding (i ) and ( ii ) ∴P_1(P_1+P_2)+P_2(P_1+P_2)=x+y P_1^2+P_1P_2+P_2P_1+P_2^2=x+y P_1^2+P_2^2+2P_1P_2=x+y ⇒(P_1+P_2)^2=x+y ⇒P_1+P_2= x+y ∴ Total pressure (P_T )=P_C+P_B+P_E (P_1+P_2)+P_1+P_2=2(P_1+P_2) P_T=2 ( x+y )

$Two solids dissociate as follows A (s) \to B (g) + C (g); K_{P_{1}} = x atm^{2} D (s) \to C (g) + E (g); K_{P_{2}} = y atm^{2} The total pressure when both the solids dissociate simultaneously is:$

Solution

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