Which would increase the rate of dissolving salt into water?

Three
A typical atom consists of three subatomic particles: protons, neutrons, and electrons (as seen in the helium atom below). Other particles exist as well, such as alpha and beta particles (which are discussed below). The Bohr model shows the three basic subatomic particles in a simple manner.

6 0

3 years ago

Consider the decomposition of a metal oxide to its elements, where M represents a generic metal. M 3 O 4 ( s ) − ⇀ ↽ − 3 M ( s )

Fiesta28 [93]

This is an incomplete question, here is a complete question.

Consider the decomposition of a metal oxide to its elements, where M represents a generic metal.

$M_3O_4(s)\rightleftharpoons 3M(s)+2O_2(g)$

Substance ΔG°f (kJ/mol)

M₃O₄ -9.50

M(s) 0

O₂(g) 0

What is the standard change in Gibbs energy for the reaction, as written, in the forward direction? delta G°rxn = kJ / mol.

What is the equilibrium constant of this reaction, as written, in the forward direction at 298 K?

What is the equilibrium pressure of O₂(g) over M(s) at 298 K?

Answer :

The Gibbs energy of reaction is, 9.50 kJ/mol

The equilibrium constant of this reaction is, 0.0216

The equilibrium pressure of O₂(g) is, 0.147 atm

Explanation :

The given chemical reaction is:

$PCl_3(l)\rightarrow PCl_3(g)$

First we have to calculate the Gibbs energy of reaction $(\Delta G^o)$ .

$\Delta G^o=G_f_{product}-G_f_{reactant}$

$\Delta G^o=[n_{M(s)}\times \Delta G^0_{(M(s))}+n_{O_2(g)}\times \Delta G^0_{(O_2(g))}]-[n_{M_3O_4(s)}\times \Delta G^0_{(M_3O_4(s))}]$

where,

$\Delta G^o$ = Gibbs energy of reaction = ?

n = number of moles

Now put all the given values in this expression, we get:

$\Delta G^o=[3mole\times (0kJ/mol)+2mole\times (0kJ/mol)]-[1mole\times (-9.50kJ/K.mol)]$

$\Delta G^o=9.50kJ/mol$

The Gibbs energy of reaction is, 9.50 kJ/mol

Now we have to calculate the equilibrium constant of this reaction.

The relation between the equilibrium constant and standard Gibbs free energy is:

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

where,

$\Delta G^o$ = standard Gibbs free energy = 9.50kJ/mol = 9500 J/mol

R = gas constant = 8.314 J/K.mol

T = temperature = 298 K

K = equilibrium constant = ?

$9500J/mol=-(8.314J/K.mol)\times (298K)\times \ln (K)$

$K=0.0216$

The equilibrium constant of this reaction is, 0.0216

Now we have to calculate the equilibrium pressure of O₂(g).

The expression of equilibrium constant is:

$K=(P_{O_2})^2$

$0.0216=(P_{O_2})^2$

$P_{O_2}=0.147atm$

The equilibrium pressure of O₂(g) is, 0.147 atm

5 0

3 years ago