If 23.4 grams of hydrogen gas were produced by the reaction below, how many grams of aluminum chloride were produced?

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

VladimirAG [237]

Answer:

For 1: The standard Gibbs free energy change of the reaction is 10.60 kJ/mol

For 2: The equilibrium constant for the given reaction at 298 K is $1.386\times 10^{-2}$

For 3: The equilibrium pressure of oxygen gas is 0.0577 atm

Explanation:

For 1:

The equation used to calculate standard Gibbs free energy change of a reaction is:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]$

For the given chemical reaction:

$M_2O_3(s)\rightarrow 2M(s)+\frac{3}{2}O_2(g)$

The equation for the standard Gibbs free energy change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(M(s))})+(\frac{3}{2}\times \Delta G^o_f_{(O_2(g))})]-[(1\times \Delta G^o_f_{(M_2O_3(s))})]$

We are given:

$\Delta G^o_f_{(M_2O_3(s))}=-10.60kJ/mol\\\Delta G^o_f_{(M(s))}=0kJ/mol\\\Delta G^o_f_{(O_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (0))+(\frac{3}{2}\times (0))]-[(1\times (-10.60))]\\\\\Delta G^o_{rxn}=10.60kJ/mol$

Hence, the standard Gibbs free energy change of the reaction is 10.60 kJ/mol

For 2:

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K_{eq}$

where,

$\Delta G^o$ = Standard Gibbs free energy = 10.60 kJ/mol = 10600 J/mol (Conversion factor: 1 kJ = 1000 J )

R = Gas constant = 8.314 J/K mol

T = temperature = 298 K

$K_{eq}$ = equilibrium constant = ?

Putting values in above equation, we get:

$10600J/mol=-(8.314J/Kmol)\times 298K\times \ln (K_{eq})\\\\K_{eq}=1.386\times 10^{-2}$

Hence, the equilibrium constant for the given reaction at 298 K is $1.386\times 10^{-2}$

For 3:

The expression of $K_{eq}$ for above equation follows:

$K_{eq}=p_{O_2}^{3/2}$

The concentration of pure solids and pure liquids are taken as 1 in the expression. That is why, the concentration of metal and metal oxide is taken as 1 in the expression.

Putting values in above expression, we get:

$1.386\times 10^{-2}=p_{O_2}^{3/2}\\\\p_{O_2}=0.0577atm$

Hence, the equilibrium pressure of oxygen gas is 0.0577 atm

6 0

3 years ago

• If 2.5 moles of sugar are added to 2.0 L of a water/lemon juice solution, what is the molarity of the lemonade?

d1i1m1o1n [39]

The answer is 4.5 because you add rhem

6 0

3 years ago

Based on position in the periodic table and electron configuration, arrange these elements in order of decreasing Ei1.

irakobra [83]

Answer:

Rb<K<Ga<As<Se<S

Explanation:

We must remember that first ionization energy decreases down the group and increases across the period.

First ionization energy decreases down the group because of the addition of more shells which increases the distance between the nucleus and the outermost electron. Hence, Rb has a lower ionization energy that K.

Across the period, increase in the size of the nuclear charge causes the pull of the nucleus on the outermost electrons to increase thereby increasing the ionization energy. Hence ionization energy increases across the period. For this reason, the ionization energy of Ga<As<Se as shown.

4 0

4 years ago

Find the molarity: 9.82 grams of lead (IV) nitrate are dissolved to make 465 mL of solution.

solmaris [256]

Molarity is moles per liter. Therefore you must covert grams->moles by dividing 9.82 by the molar mass of lead IV nitrate. Then convert mL->L. Once you have those two numbers you can plug them into the formula Molarity = mol/L.

3 0

3 years ago

Calculate the mass (in g) of 2.8 X 10^22 H2O molecules.

siniylev [52]

Answer: wait im gonna search some info.

Explanation:I come back with an asnwer for you

6 0

3 years ago