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Question

$${ K }_{ c }$$ for the reaction ; $${ N }_{ 2 }+3{ H }_{ 2 }\rightleftharpoons 2{ NH }_{ 3 }$$ is 0.5 $${ mol }^{ -2 }L^{ 2 }$$ at 400 K. Find $${ K }_{ P }$$.
[Given that $$R = 0.082 \ L-atmK^{-1}{ mol }^{ -1 }$$]


A
KP=4.648×104(atm)2
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B
KP=46.36×105(atm)2
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C
KP=8.12×105(atm)2
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D
KP=2.190×103(atm)2
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Solution

The correct options are
A $${ K }_{ P }=4.648\times { 10 }^{ -4 }{ (atm) }^{ -2 }$$
B $${ K }_{ P }=46.36\times { 10 }^{ -5 }{ (atm) }^{ -2 }$$
Given that;

$${ N }_{ 2 }+3{ H }_{ 2 }\rightleftharpoons 2{ NH }_{ 3 }$$

$$K_c=0.5\ M^{-2}$$

$$\Delta n_g =2-(1+3)=-2$$

$${ K }_{ P }={ K }_{ C }{ (RT) }^{ \triangle n_g }$$
$$\therefore { K }_{ P }=0.5\times { (0.082\times 400) }^{ -2 }$$
$${ K }_{ P }=4.648\times { 10 }^{ -4 }{ (atm) }^{ -2 }$$

Chemistry

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