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The half life for the reaction N_(2)O_(5) hArr 2NO_(2)+1/2 O_(2) in 24 hr at 30^(@)C. Starting with 10 g of N_(2)O_(5) how many grams of N_(2)O_(5) will remain after a period of 96 hours ? |
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Answer» Solution :`K = (0.063)/(24) hr^(-1) = (2.303)/(96)` log `(10)/(a -X)` or log `(10)/(a-x) = 1.2036` or 1 - log ( a-x) = `1.2036` or log (a-x) = `-0.2036 = 1.7964` or (a-x) = antilog 1.7964 or (a-x) = antilog 1.7964 = 0.6258 GM . |
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