If a reaction is second-order with respect to a particular reactant, when the concentration of that reactant is increased by a factor of 4, the reaction rate will increased by factor 16.
<h3>How do we calculate the rate of second order reaction?</h3>
Rate of the second order reaction will depends on the concentrations of one second-order reactant or two first-order reactants as:
Rate(1) = k[A]²
If the concentration of reactant becomes 4 times the initial concentration then rate will be:
Rate = k[4A]²
Rate(2) = 16k[A]²
Rate(2) = 16Rate(1)
Hence rate will increased by 16 times.
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Not at all they’re poisonous
Aldehydes and ketones having
α-hydrogen atoms, undergoes aldol condensation, in present of base (NaOH).
The initial product formed during this reaction is
β-hydroxy alcohol, which then undergoes dehydration to form
α,β-unsaturated aldehyde or ketone.
In present case, 3,3-dimethyl-2-butanone has 2α-hyrogen atom, while methylcyclopentane-1-carbaldehyde has 1α-hydrogen atom. So the major product formed during cross aldol condensation reaction of these reactants is:
5-hydroxy-4,4-dimethly-1-(2-methylcyclopentyl)pent-1-en-3-one. The complete reaction product formed is shown below.
Balanced equation:
Mg + 2 HNO3 —> Mg(NO3)2 + H2
This is a metal + acid reaction giving salt and hydrogen (not water).