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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 ?A. `1.25 g`B. `0.63 g`C. `1.7 g`D. `0.5 g` |
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Answer» Correct Answer - B `k = (0.693)/(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) = "anitlog" 1.7964 = 0.6258g` |
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