The thermal decomposition of acetaldehyde : CH 3 CHO → CH 4 + CO, has rate constant of…
Chemistry · JEE Advanced · NTA Exams — Chemical Kinetics
The thermal decomposition of acetaldehyde :
CH3CHO\(\text { → }\)CH4 + CO, has rate constant of 1.8 × 10–3 mole–1/2 L1/2 min–1 at a given temperature. How would \(-\frac{\mathrm{d}\left[\mathrm{H}_{3} \mathrm{CHO}\right]}{\mathrm{t}}\) will change if concentration of acetaldehyde is doubled keeping the temperature constant ?
CH3CHO\(\text { → }\)CH4 + CO, has rate constant of 1.8 × 10–3 mole–1/2 L1/2 min–1 at a given temperature. How would \(-\frac{\mathrm{d}\left[\mathrm{H}_{3} \mathrm{CHO}\right]}{\mathrm{t}}\) will change if concentration of acetaldehyde is doubled keeping the temperature constant ?
- will increase by 2.828 times
- will increase by 11.313 times
- will not change
- will increase by 4 times
Answer
(A) will increase by 2.828 times
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