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Question

The reaction between N2 and H2 to form ammonia has KC=6×102 at the temperature 500oC. The numerical value of Kp for this reaction is;

A
1.5×105
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B
1.5×105
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C
1.5×106
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D
1.5×106
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Solution

The correct option is A 1.5×105
N2+3H22NH3

Given : KC=6.0×102:T=500C=500+273=773K

Δn=231=2

Kp=Kc(RT)Δn=6×102×(0.0821×773)2

=6×102×(63.46)2=6×102(63.46)2=6×1024027.17=0.00148×102

Kp=1.5×102×103=1.5×105

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