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The half life for the reaction `N_(2)O_(5) hArr 2NO_(2) + (1)/(2) O_(2)` is 24 hrs 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 hoursA. 1.25 gB. 0.63 gC. 1.77 gD. 0.5 g |
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Answer» Correct Answer - b `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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