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Question

Two solid compounds $$X$$ and $$Y$$ dissociates at a certain temperature as follows:
$$X(s)\rightleftharpoons A(g)+2B(g); { K }_{ { P }_{ 1 } }={ 9\times 10 }^{ -3 }{ atm }^{ 3 }$$
$$Y(s)\rightleftharpoons 2B(g)+C(g); { K }_{ { P }_{ 2 } }={ 4.5\times 10 }^{ -3 }{ atm }^{ 3 }$$
The total pressure of gases over a mixture of $$X$$ and $$Y$$ is:


A
4.5 atm
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B
0.45 atm
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C
0.6 atm
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D
None of these
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Solution

The correct option is B $$0.45$$ atm
$$X(s)\rightleftharpoons A(g)+2B(g); { K }_{ { P }_{ 1 } }={ 9\times 10 }^{ -3 }{ atm }^{ 3 }$$  ......(1)
$$Y(s)\rightleftharpoons 2B(g)+C(g); { K }_{ { P }_{ 2 } }={ 4.5\times 10 }^{ -3 }{ atm }^{ 3 }$$ ......(2)
Add reactions (1) and (2)
$$X(s) + Y(s)\rightleftharpoons  A(g)+4B(g)+C(g) $$
$${ K }_{ { P }}= { K }_{ { P }_{ 1 } } \times { K }_{ { P }_{ 2 } }={ 9\times 10 }^{ -3 }{ atm }^{ 3 }\times { 4.5\times 10 }^{ -3 }{ atm }^{ 3 }$$
Let P be the pressure of A.
The pressures of B and C will be 4P and P respectively.
Hence, the expression for the equilibrium constant will be $$K_P=P_{A}\times P_{B}^4 \times P_{C}$$
Substitute values in the above expression.
$$P \times (4P)^4 \times P = 9 \times 4.5 \times 10^{-3} \times 10^{-3}$$
$$P=0.0735$$
$$6P=6 \times 0.0735 = 0.45 \: atm$$
Hence, the total pressure of gases over a mixture of X and Y is 0.45 atm.

Chemistry

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