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2 years ago

Consider the reaction and the rate law below.

Chemistry

1 answer:

ruslelena [56]2 years ago

3 0

The overall order of the reactants in this rate law : 2

<h3>Further explanation</h3>

Given

Reaction

2O3 → 302

The rate law :

R= k[03]²

Required

The overall order

Solution

The overall order : the sum of orders for each reactant

For A + B ---> C + D

r = k [A]ᵃ[B]ᵇ

The overall reaction order is m+n

So for the rate law of R= k[03]², the overall order = 2

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Calculate the cell potential, E, for the following reactions at 26.29 °C using the ion concentrations provided. Then, determine

yanalaym [24]

Answer:

I will work only one of the listed equations ... you follow the given example for the remaining reactions. Thank you :-)

Rxn 1: Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s)

a) E(Pt⁺²/Fe°) = - 1.668v

b) Process is Non-spontaneous if E(cell) < 0

Explanation:

Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s) ⇔

Pt°(s)|Pt⁺²[0.057M]║Fe⁺²[0.006M]|Fe°(s)

As written, Pt° is shown undergoing oxidation with Fe⁺² undergoing reduction. Applying the reduction potentials to the analytical equations for E(cell) and ΔG(cell) gives E(Pt/Fe⁺²) < 0 and ΔG(Pt/Fe⁺²) > 0 which indicate a non-spontaneous process. The following supports this conclusion.

E°(Fe⁺²) = -0.44v

E°(Pt⁺²) = +1.20v

E°(Pt/Fe⁺²) =E°(Redn) - E°(Oxidn) =E°(Fe⁺²) - E°(Pt⁺²)

= -0.44v - (+1.20v) = - 1.64v

[Fe⁺²] = 0.0066M

[Pt⁺²] = 0.057M

n = electrons transferred = 2

E(nonstd) = E°(std) - (0.0592/n)logQ);

Q = [Pt⁺²]/[Fe⁺²]

= -1.64v - (0.0592/2)log[0.057M]/[0.006M]v = -1.668v

Also, if ΔG(cell) > 0 => indicates non-spontaneous process

ΔG(Pt/Fe⁺²) = - nFE = -(2)(96,500Coulombs)((-1.664v) > 0 Kj => nonspontaneous rxn. (1 Coulomb-volt = 1 Kilojoule)

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3 years ago

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yarga [219]

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Have a nice day!

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3 years ago

What is the mass of a sample of alcohol (specific heat = 2.4 J/gC), if it requires 4780 J of heat to raise the temperature by 5.

insens350 [35]

The mass of a sample of alcohol is found to be = m = 367 g

Hence, it is found out that by raising the temperature of the given product, the mass of alcohol would be 367 g.

Explanation:

The Energy of the sample given is q = 4780

We are required to find the mass of alcohol m = ?

Given that,

The specific heat given is represented by = c = 2.4 J/gC

The temperature given is ΔT = 5.43° C

The mass of sample of alcohol can be found as follows,

The formula is c = $\frac{q}{mt}$

We can drive value of m bu shifting m on the left hand side,

m = $\frac{q}{ct}$

mass of alcohol (m) = $\frac{4780}{(2.4)( 5.43)}$

m = 367 g

Therefore, The mass of the given sample of alcohol is

m = 367g

It requires 4780 J of heat to raise the temperature by 5.43 C in the process which yields a mass of 367 g of alcohol.

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3 years ago