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In a zero-order reaction for every `10^(@)` rise of temperature, the rate is doubled. If the temperature is increased from `10^(@)C` to `100^(@)C`, the rate of the reaction will becomeA. `64` timesB. `512` timesC. `256` timesD. `128` times |
Answer» Correct Answer - B `(r + (r +100))/(r_(t)) = 2` for each `10^(@)` rise in temperature `:. (r_(100))/(r_(10)) = (2)^(9) = 512` |
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