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For the reactions N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g), " at " 400 K, K_(p) = 41." Find thevalue of "K_(p) for each of the followingreactions at the same temperature : (i)2 NH_(3) (g) hArr N_(2) (g) + 3 H_(2) (g)(ii) 1/2 N_(2) (g) + 3/2H_(2) (g) hArr NH_(3) (g)(iii) 2 N_(2) (g) + 6 H_(2) (g) hArr 4 NH_(3) (g) |
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Answer» Solution :(i) It is the reverse of the GIVEN reaction. Hence , `K_(p) = 1/41` (ii) It is obtained by DIVIDING the given equation by 2. Hence, `K_(p) = sqrt(41)` (iii) It is obtained by MULTIPLYING the given equation by 2. Hence, `K_(p) = (41)^(2)`. |
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