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1

The partial reaction half-life relates to the reaction speed constant by the same way as for the "normal" reaction half-life. If there are 2 parallel reactions of the first order, $\ce{A -> B}$ and $\ce{A -> C}$, and if there is the reaction rate for the former: $$\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t}=k_{\ce{B}} \cdot [\ce{A}]$$ then the ...

0

The rate has units $\displaystyle \frac{concentration}{time}$ and is $\displaystyle \frac{d[C]}{dt}$ and the right hand side of the equation has always to have the same units as the left hand side, so for first order reaction consuming C; $\displaystyle \frac{d[C]}{dt} = -k_1[C]$ and the rate constant $k_1$ therefore has units $\displaystyle \frac{1}{time}$ ...

0

By the Arrhenius equation, $k$ (rate constant) can be expressed as, $$k = \rho Z \times \exp\bigg(-\frac{E_a}{RT}\bigg)$$ where $\rho$ (aka. steric factor, or probability factor) is defined as the ratio of $A$ (Arrhenius contant or pre-exponential factor) and $Z$ (collision frequency) (Ref. 1). And, units of $A$ are same as that of units of $k$ (Ref. 2), ...

5

The reason might be that while drawing the reaction energy profile, we forget to mention what energy we are mentioning in the Y-axis. The following conventions are generally used: If reaction conditions are constant NVT, energy in Y-axis should represent internal energy. If reaction conditions are constant NPT, energy in Y-axis should represent enthalpy. If ...

3

If the reaction is $$\ce{2A -> P},$$ then the rates are defined to be related as $$-\frac{1}{2}\frac{\mathrm{d}\ce{A}}{\mathrm{d}t} = \frac{\mathrm{d}\ce{P}}{\mathrm{d}t}$$ and this is true throughout the reaction. The rate at which $\ce{P}$ is produced is half that at which $\ce{A}$ is consumed. Normally therefore a factor of 2 is expected with ...

1

The first expression $$\frac{\mathrm{d}[A]}{\mathrm{d}t} = -2k_\mathrm{f}[\ce{A}]^{2}\tag{1}$$ is the correct one for the reaction given. As the OP mentions in the comments, both approaches describe the situation correctly. However, the values of $k_\mathrm{f}$ would be different by a factor of two. You could rewrite the chemical equation by dividing by two (...

5

As the system is described, we can suppose $k_1 \ll k_2$ ans $k1 \ll k_3$, as Martin correctly noted and I have omitted to explicitly mention. For such cases, we can consider for intermediate products to be in a steady state, i.e. $\frac {\mathrm{d}[A]}{\mathrm{d}t} \simeq 0$. So the rate of their creation is about equal to the rate of there destruction. E.g....

0

The whole reaction isn't first order and that can be arrived at using one simple concept. The unit of rate constant $k$ is $\pu{M^{-1} s^{-1}}$ and not $\mathrm s^{-1}$. Since reaction is first order with respect to both reactant A and B, I can write, $$k=\frac{1}{t} \ln{\frac{a}{a-x}}$$ This would have been correct if and only if the whole reaction was ...

2

The rate-determining step of halogenation of alkenes is the formation of cyclic intermediate. The cyclohalonium intermediate of bromine will be more stable than that of chlorine owing to its lower electronegativity. The following paragraph is taken from Peter Sykes [$1$, p.$181$-$182$]: It is not normally possible to add fluorine directly to alkenes as the ...

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