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3

You have not propagated your errors properly. Note that the sum in your nominator evaluates to a small number with an error of roughly the same size (99.73±0.01 - 99.72±0.01 = 0.01±0.01). Also, error propagation cannot be done by simply adding up the errors. Instead of calculating the error propagation manually (cumbersome, and prone to errors, hehe ;-) I ...


0

I think what OP need is the derivation of the equation $(1)$ in the question (as indicated in the OP's comment, "an anyone help me with the derivation of the first one"). The first part of this derivation, the formula for the rate constant $k$ for the first order reaction, is done in the last part of Chemistry student's answer, and hence I'd not ...


1

The half-life $t_½$ is the time it takes for the concentration of the first order reaction to decrease by a factor of two. If we plug this into equation (2) given by the OP, we get $$\ln\frac{1}{2} = -k t_½$$ We solve this for $k$ to get: $$k = \frac{ \ln(2) }{t_½} \approx \frac{0.693}{t_½} $$ My lecturer mentioned that the formula for the rate constant $k$ ...


3

I am attaching a written answer as typing the log and other exponents is hard :(


1

The two equations are pretty the same. The number 2.30 is used to convert the natural logarithm in decimal logarithm: $\ln 10 = 2.303$. Also, substitute $[A]_{1/2} = [A]_0 /2$, from the definition. The concentration at the half-life is exactly the half of the initial concentration, then you will obtain $\ln 2$ inside the logarithm term.


3

Well, first of all, I should point out that your answer and the would be the answer ($1.7$ times your answer) are both wrong by the fact that after 552 days (4 half-lives), the activity should be $\displaystyle\frac{\pu{2E6 MBq}}{16} = \pu{1.25E5 MBq}.$ Since this is smaller than both your answer and the would be answer, they both would be incorrect. ...


4

By saying Catalysts are not consumed by reactions. is meant there is no stoichimetric ratio to reactants, consuming catalysts and forming from them catalytically inactive compound. By other words, some of reaction steps regenerates the original form of a catalyst, consumed by a prior step, so the net consumption is negligible. The transition between active ...


1

I'll try and answer your question, but a "full" answer would take a book. Given the reaction: $$\ce{aA + bB <=> cC + dD}$$ Then assuming an elementary reaction in the gaseous state the concentration equilibrium constant always has products over reactants and will be: $$K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}$$ The above equation relies on two ...


1

Let's first assign rate constants to each reaction for completeness: $\ce{2X + Y <=>[k_1][k_{-1}] 3X}$ $\ce{A ->[k_2] Y}$ $\ce{X->[k_3] B}$ Then let's consider what's happening overall. A is converting to Y, Y is converting to X, and X is converting to B: $\ce{A ->[k_2] Y <=>[k_1X^2][k_{-1}X^2] X ->[k_3] B}$ Y and X may be in ...


1

Let's take a numerical example: $\ce{3 A + 1 B -> 8 C + 2 D}$ Here : $a = -3, b = -1, c = 8, d = 2$. Let's start from $100$ $mol$ of $\ce{A}$, $120$ $mol$ of $\ce{B}$, $1$ $mol$ $\ce{C}$, and no $\ce{D}$. Suppose that $10$ $mol$ $\ce{B}$ have reacted with $30$ $mol$ $\ce{A}$, and producing $80$ $mol$ $\ce{C}$ and $20$ $mol$ $\ce{D}$. At the end of the ...


1

Yes, your expression is correct. It's nothing new though. Entropy of activation can be interpreted as part of the pre-exponential factor in the Eyring equation: $$k=\bigg(\frac{k_B T}{h}e^{\frac{\Delta^\ddagger S^\circ}{R}}\bigg)\cdot e^{-\frac{\Delta^\ddagger H^\circ}{RT}}$$


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