Please assist me with chemistry assessment

2.15 Liters of helium gas at a pressure of 58 kPa and a temperature of 25 degrees celsius, what is the new volume at stp?

Nastasia [14]

The ideal gas law is P1V1/T1 = P2V2/T2. STP means the temperature is 273 K and pressure is 101.3 kPa. According to this formula, the new volume V2=2.15*58*273/(298*101.3) = 1.13 L.

5 0

4 years ago

Indicate whether the following interactions of matter include a physical or chemical change. P/C? Evidence

Anarel [89]

1. CHEM CHANGE
2. PHYS CHANGE
3. CHEM CHANGE
4. CHEM CHANGE
5. PHYS CHANGE

3 0

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:

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