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The equilibirum constant `(K_(c))` of two reactions `H_(2)+I_(2)hArr2HIand N_(2)+3H_(2)hArr2NH_(3)` are 50 and 1000, respectively. The equilibirum constant of the raction `N_(2)4=6HI2NH_(3)+3I_(2)` is closeest to :A. 50000B. 20C. `0.008`D. `0.005` |
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Answer» Correct Answer - C `underline({:(H_(2)+I_(2)hArr2HI, ,K_(c)=50),(N_(2)3H_(2)hArr2NH_(3),,K_(c)=100):})` `{:(N_(2)+6HIhArr2NH_(3)+3I_(2),,K_(c)=(1000)/((50)^(3))),(,K_(c)=0.005):}` |
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