Extra Credit 35

12.5.7

How much faster does the reaction proceed at \(45^\circ\) then \(25^\circ\) degrees? How much faster does the reaction proceed at \(95^\circ\) then \(25^\circ\)?

S12.5.7

The rate doubles for every \(10^\circ\) rise, so if we start at \(25^\circ\) degrees then go to \(45^\circ\), the reaction has gone up \(25^\circ\) so it has doubled twice. $$2\times2=4x$$ so the reaction goes 4 times faster then what we originally started with.

For the reaction going from \(95^\circ\) to \(25^\circ\)

So since the reactions goes from \(25^\circ\) to \(95^\circ\), so lets go through in increments of \(10^\circ\) to find out how much it doubles $$25\, to \,35 = 2x\,rate$$ $$35\, to \,45 =4x\,rate$$ $$45\,to\, 55= 8x\,rate$$ $$ 55\,to\,65=16x\, rate$$ $$65\,to\,75=32x\, rate$$ $$75\,to\, 85=64x\, rate$$ $$85\,to \,95=128\,rate$$ so it doubles 7 times for every \(10^\circ\) and is 128 times faster at \(95^\circ\) then \(25^\circ\)

Edit: This question is correct. In general, \(25^\circ\) is the standard temperature in Celsius. It can be converted to 298 Kelvin.

Q21.4.2

What changes occur to the atomic number and mass of a nucleus during each of the following decay scenarios?

1. an \(\alpha\) particle is emitted
2. a \(\beta\) particle is emitted
3. \(\gamma\) radiation is emitted
4. a positron is emitted
5. an electron is captured

S21.4.2

A alpha particle consists of two protons and two neutrons bound together into a particle identical to a helium nucleus. When alpha particle is emitted: A 2 protons will be lost and mass will decrease by 4 $$_2^4\alpha$$ $$Example: _{106}^{263}Sg\rightarrow_{104}^{259}Rf+_2^4He$$

Beta decay occurs when a proton or neutron is transformed into the other because there are to many in the nucleus. There is beta minus decay and beta plus decay. In beta minus decay a neutron decays into a proton, electron and antineutrino. When a beta particle is emitted: The proton will be gained and the mass will not be effected $$_{-1}^0\beta$$$$Example:\,_{29}^{61}Cu\rightarrow_{30}^{61}Zn+_{-1}^0\beta$$

When gamma radiation is emitted: The mass and proton are not effected $$_0^0\gamma$$

A positron is a subtype of beta emission. When a positron is emitted: A proton is lost and mass is not effected $$_1^0\beta$$$$Example:\,_{25}^{50}Mn\rightarrow_{24}^{50}Cr+_1^0e$$

During electron capture the electron in the atoms most inner shell is pulled toward the nucleus and combines with a proton. This forms a neutron and a neutrino which is ejected from the nucleus. When a electron is captured: A proton is lost and the mass is not effected $$_1^0e$$ $$Example:\,_{30}^{63}+_{-1}^0e\rightarrow_{29}^{63}Cu$$

Edit: The provided explanations are correct.

20.2.6

Of these elements, which would you expect to be easiest to reduce: \(Se\), \(Sr\), or \(Ni\)? Explain your reasoning.

S20.2.6

Out of the elements \(Se\), \(Sr\), and \(Ni\), \(Sr\) is the easiest to reduce. Based on Periodic Trends, ionization energy increases as you go to the right and decreases as you go down (increases as you go up). Ionization energy is the energy required to remove an electron from a atom. Reduction is related to Ionization energy and has the same trends. So based on that Periodic Trend, \(Se\) is the most far right element out of the 3 and therefore the easiest to reduce.

Edit: There are different "levels" of ionization energies. The first ionization energy is the energy needed to remove the outermost electron from a neutral atom in the gas phase. When releasing energy, the removal of the following electrons after the first, leads to less energy being released each time.

Q20.5.1

State whether you agree or disagree with this reasoning and explain your answer: Standard electrode potentials arise from the number of electrons transferred. The greater the number of electrons transferred, the greater the measured potential difference. If 1 mol of a substance produces 0.76 V when 2 mol of electrons are transferred—as in \(Zn(s)\rightarrow\, Zn^2+(aq) + 2e^-\,\)—then 0.5 mol of the substance will produce 0.76/2 V because only 1 mol of electrons is transferred.

S20.5.1

I would disagree because standard reduction potential is independent of size and amount of matter in a system. So, the standard reduction potential remains the same for a given metal-metal ion couple. It doesn't matter if there was one electron transfer or two electrons transferred. The standard reduction potential of $$Zn(s)\rightarrow\,Zn^2+2e^-$$ is the same.

Edit: I disagree as well because the quantity of electrons does not cause an effect. The term "standard" can be applied to standard conditions.

Q20.9.10

Electrolysis of \(Cr^{3+}\) produces \(Cr^{2+}\). If we had 500mL of a 0.15M solution of \(Cr^{3+}\) how long would it take to reduce the \(Cr^{3+}\) to \(Cr^{2+}\) using a 0.158A current?

S.20.9.10

Start by writing the equation and formula you will need

$$Cr^{3+}+e^-\rightarrow\,Cr^{2+}$$

$$m=\frac{IT}F \cdot\frac{M}{n}$$

n= number of moles transferred

F= Faradays constant

M= molarity

I=current

t=time

That equation can be reduced to \(Q=IT\)

First we find n, so we are going from \(Cr^{3+}\) to \(Cr^{2+}\), so there is only one electron that is transferred.

$$n=1mole\,e^-\, (going\, from \,Cr^{3+}\, to \,Cr^{2+})$$

Next, we need to get volume into Liters to isolate moles from the molarity

$$Volume=500mL\times\frac{10^-3L}{1mL}$$

$$Volume=0.5L$$

$$moles\, of\, Cr^{3+}=.15\frac{mol}{L}\times0.5L$$

$$moles\, of\, Cr^{3+}=0.075mol$$

Now that we have moles, we can use the number of moles, faradays constant \(F\), and the number of electrons transferred \(n\) to find \(Q\)

$$F=96487\frac{C}{mol\,e^-}$$

$$F=96487\frac{C}{mol\,e^-}\times1mol\,e^-$$

$$F=96487C$$

$$Q=0.075mol Cr^3\times96487{C}=7236.5C$$

Now we can plug everything into the equation. Don't forget to convert \(C\) to \(A\) conversion \(1A=1C\)

$$Q=It$$

$$t=\frac{Q}{I}$$

$$t=\frac{7236.5C}{0.158A}\times\frac{A}{C}$$

$$t=45800.8s$$

So it would have taken 45800.8 seconds

Edit: Everything for this question appears to be correct.

Q14.6.8

Nitramide \(O_2NNH_2\) decomposes in aqueous solution to \(N_2O\) and \(H_2O\). What is the experimental rate law \(\frac{\nabla[N_2O]}{\nabla\,t}\) for the decomposition of nitramide if the mechanism for the decomposition is as follows? $$O_2NNH_2\Leftrightarrow^{k_1}_{k_{-1}}\,O_2NNH^-+H^+(fast)$$

$$O_2NNH^-\rightarrow^{k_2}\,N_2O+OH^-(slow)$$

$$H^++OH^-\rightarrow^{k_3}\,H2O(fast)$$

S14.6.8

So, the rate for the overall reaction is equal to the rate law for the slow step $$rate=rate_2=k_2[O_2NNH^-]$$

But, Nitramide is a intermediate so we have to substitute it for the equivalent expression from the equilibrium step.

$$rate_1=rate_{-1}$$

For the first fast step the reaction is going forward and reverse so \(k_1\, and\, k_{-1}\) rates equal each other

$$k_1[O_2NNH_2]=k_{-1}[O_2NNH^-][H^+]$$

Now we want to get \(O_2NNH^-\) alone because it's the intermediate

$$[O_2NNH^-]=\frac{k_1[O_2NNH_2]}{k_{-1}[H^+]}$$

The second reaction in the mechanism is the slow step and the rate of the reaction depends on the slow step. So

$$rate=rate_2=k_2[O_2NNH^-]$$

Now we can just plug in \(O_2NNH^-\) into the slow step to get the overall rate of reaction

$$rate=k_2\times\frac{k_1[O_2NNH_2]}{k_{-1}[H^+]}$$

In terms of just k, the rate would be

$$rate=k=\frac{k_1k_2}{k_{-1}}$$

The overall rate of the reaction is

$$rate=k\frac{[O_2NNH_2]}{[H^+]}$$

Edit: Notice that the overall rate of a reaction doesn't include the intermediate product in the final rate law answer. Intermediates can be isolated because of their fully formed bonds.