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A substance A decomposes by a first order reaction starting initially with [A] = 2.00 m and after 200 min [A] = 0.15 m . For this reaction what is the value of kA. `1.29 xx 10^(-2) "min"^(-1)`B. `2.29 xx 10^(-2) "min"^(-1)`C. `3.29 xx 10^(-2) "min"^(-1)`D. `4.40 xx 10^(-2) "min"^(-1)` |
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Answer» Correct Answer - a Given A(a) = `2.00` m, t = 200 min and a(a-x) = 0.15 m we know `k = (2.303)/(t) "log" (a)/(a-x) = (2.303)/(200) "log" (2.00)/(0.15)` `= (2.303)/(200) xx (0.301 + 0.824) = 1.29 xx 10^(-2) "min"^(-1).` |
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