Using standard electrode potentials calculate δg∘rxn and use its value to estimate the equilibrium constant for each of the reactions at 25 ∘c.

          K = 1.26 X10^-10

Explanation:

We have the formula,

By the Nernst equation:

E = Eo - (0.0592/n)(log Q)

Put E=0which at equilibrium becomes

0 = Eo - (0.0592/n) (log K)

which re-write the equation as

log K = Eo / (0.0592/n)

1.Pb2+(aq)+Mg(s)->Pb(s) + Mg2+ (aq)

Finding the value of Eo we get,

Pb2+(aq)--> Pb(s) Eo = –0.125

Mg(s) -> Mg2+ (aq) Eo = + 2.356

Pb2+(aq)+Mg(s)->Pb(s) + Mg2+ (aq) Eo = 2.231

Subsitute in this formula we get

log K = Eo / (0.0592/n)

log K = 2.231 / (0.0592/2)

log K = 75.37

K = 2.35 X10^75

The value of K for the equation Pb2+(aq)+Mg(s)->Pb(s) + Mg2+ (aq) is K = 2.35 X10^75

2.Br2 (l)+2Cl-(aq)->2Br-(aq)+Cl2 (g) Eo = -0.293

Similarly, for the other equation  follow the same procedure

The equation is written in the form of non-spontaneous

2Cl-(aq) -> Cl2 Eo = -1.358

Br2 (l) --> 2Br-(aq Eo = 1.065

Br2 (l)+2Cl-(aq)->2Br-(aq)+Cl2 (g) Eo = -0.293

log K = -0.293 / (0.0592/2)

log K = - 9.90

K = 1.26 X10^-10

The value of k for the equation Br2 (l)+2Cl-(aq)->2Br-(aq)+Cl2 (g)  is

K = 1.26 X10^-10

3.MnO2 (s) +4H+ (aq) + Cu (s) -> Mn2+ (aq) +2H2O(l) + Cu2+(aq)

Similarly, for the other equation  follow the same procedure

The equation is written in the form of non-spontaneous

MnO2 (s) -> Mn2+ Eo = 1.23

Cu (s) -> Cu2+ Eo = - 0.340

MnO2 (s) +4H+ (aq) + Cu (s) -> Mn2+ (aq) +2H2O(l) + Cu2+(aq) Eo = 0.89

log K = 0.89 / (0.0592/2)

log K = 30.067

K = 1.17 X10^30

The value of k for the equation MnO2 (s) +4H+ (aq) + Cu (s) -> Mn2+ (aq) +2H2O(l) + Cu2+(aq) Eo is K = 1.17 X10^30

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j