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3 years ago

I need help help me please

Chemistry

1 answer:

Rzqust [24]3 years ago

3 0

Answer:

C is the correct answer.

Explanation:

We can eliminate some answers immediately:

A is incorrect because nitrogen and hydrogen do not change their identity, they just combine into a new molecule that contains them. Notice how there are 2 nitrogen atoms and 6 hydrogen atoms on both sides of the equation.

B is incorrect because the first law of thermodynamics tells us that matter and energy will always be conserved, so the mass of the products must be equal to the mass of the reactants, not less than the mass of the reactants.

C is correct because balanced chemical equations help chemists predict how much product will form from certain amounts of reactants.

D is incorrect because this equation is showing the exact opposite. Molecular hydrogen and nitrogen can combine stoichiometrically to form ammonia.

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3 years ago

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:

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