In the following reaction: Mg + 2HCl → MgCl2 + H2 How many liters of H2 would be produced if you started with 24.3 g of Mg?

Silver occurs in trace amounts in some ores of lead, and lead can displace silver from solution: Pb(s) + 2Ag+ (aq) LaTeX: \longr

VikaD [51]

Answer : The value of $\Delta G^o$ and K is, -180 kJ/mol and $3.6\times 10^{31}$

Explanation :

The balanced cell reaction will be,

$Pb(s)+2Ag^+(aq)\rightarrow Pb^{2+}(aq)+2Ag(g)$

The half-cell reactions are:

Oxidation reaction (anode) : $Pb(s)\rightarrow Pb^{2+}(aq)+2e^-$

Reduction reaction (cathode) : $2Ag^+(aq)+2e^-\rightarrow 2Ag(g)$

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$

where,

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

F = Faraday constant = 96500 C

n = number of electrons in oxidation-reduction reaction = 2

$E^o_{cell}$ = standard electrode potential of the cell = 0.93 V

Now put all the given values in the above formula, we get:

$\Delta G^o=-2\times 96500\times 0.93$

$\Delta G^o=-179490J/mol=-179.49kJ/mol\approx -180kJ/mol$

Now we have to calculate the value of 'K'.

$\Delta G^o=-RT\ln K$

where,

$\Delta G_^o$ = standard Gibbs free energy = -180 kJ/mol

R = gas constant = $8.314\times 10^{-3}kJ/mole.K$

T = temperature = 298 K

K = equilibrium constant = ?

Now put all the given values in the above formula 1, we get:

$-180kJ/mol=-(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln K$

$K=3.6\times 10^{31}$

Therefore, the value of $\Delta G^o$ and K is, -180 kJ/mol and $3.6\times 10^{31}$

5 0

3 years ago

(Please help, ASAP) How many grams are in 3.45x10^23 atoms of P?

Daniel [21]

one mole of P weights about 31 grams

in one mole there are 6.022*10^23 atoms

we use the rule of threes

6.022*10^23atoms......weight..........31 grams

3.45*10^23 atoms.........weight...........x grams

x=(3.45*10^23*31)/6.022*10^23

x=106.95/6.022=17.76 grams

5 0

3 years ago

True or False? Nuclear fusion releases less energy than nuclear fission does.

elixir [45]

False

nuclear fusion produces more energy than a nuclear fission reaction.

8 0

3 years ago

Read 2 more answers

Can anyone can answer practice problem exercise 6.5

NemiM [27]

Answer: Add more information please! Thanks!

Explanation:

6 0

3 years ago

Read 2 more answers

calculate δg o for each reaction using δg o f values: (a) h2(g) i2(s) → 2hi(g) 2.6 kj (b) mno2(s) 2co(g) → mn(s) 2co2(g) kj (c)

Reika [66]

(a)The change in Gibbs free energy for the reaction has been 2.6 kJ/mol.

(b) The change in Gibbs free energy for the reaction has been -49.3 kJ/mol.

(c) The change in Gibbs free energy for the reaction has been 91.38 kJ/mol.

6 0

2 years ago