Chemistry, 10.12.2019 01:31 adajadavis2843

Consider the following reaction. mg(s) + 2vo2+ (aq)+ 4 h+(aq) → mg2+(aq) + 2 vo2+(aq) + h2o(l)e⁰cell = 3.37 v. a.calculate the δg⁰∘ for the reaction. express your answer to three significant figures. b. calculate the k for the reaction. express your answer to three significant figures.

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Answer from: cexe468

Answer :

(a) The value of Gibbs free energy for the reaction is $-6.50\times 10^5J/mol$

(b) The value of at 298 is, $8.28\times 10^{113}$

Explanation :

The given cell reaction is:

$Mg(g)+2VO_2^{+}(aq)+4H^+(aq)\rightarrow Mg^{2+}(aq)+2VO^{2+}(aq)+2H_2O(l)$

The half reaction will be:

Reaction at anode (oxidation) : $Mg\rightarrow Mg^{2+}+2e^-$

Reaction at cathode (reduction) : $2VO_2^{+}+4H^++2e^-\rightarrow 2VO^{2+}+2H_2O$

First we have to calculate the Gibbs free energy.

Formula used :

$\Delta G^o=-nFE^o$

where,

$\Delta G^o$ = Gibbs free energy = ?

n = number of electrons = 2 (from the reaction)

F = Faraday constant = 96500 C/mole

E^o = standard e.m.f of cell = 3.37 V

Now put all the given values in this formula, we get the Gibbs free energy.

$\Delta G^o=-(2\times 96500\times 3.37)=-650410J/mol=-6.50\times 10^5J/mol$

Thus, the value of Gibbs free energy for the reaction is $-6.50\times 10^5J/mol$

Now we have to calculate the value of equilibrium constant.

Formula used :

$\Delta G^o=-2.303\times RT\times \log K$

where,

R = universal gas constant = 8.314 J/K/mole

T = temperature = 25^oC=273+25=298K

K = equilibrium constant = ?

Now put all the given values in this formula, we get the value of

$-6.50\times 10^5J/mole=-2.303\times (8.314 J/K/mole)\times (298K)\times \log K$

$\log K=113.918$

$K=8.28\times 10^{113}$

Therefore, the value of at 298 is, $8.28\times 10^{113}$

Answer from: Quest

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Answer from: Quest

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.

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Consider the following reaction. mg(s) + 2vo2+ (aq)+ 4 h+(aq) → mg2+(aq) + 2 vo2+(aq) + h2o(l)e⁰cell...