1 Point

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

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

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

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

$\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]$

$\Delta S^o=-153J/K$

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

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 500 K.

$\Delta G^o=(-1035600J)-(500K\times -153J/K)$

$\Delta G^o=-959100J=-959.1kJ$

Therefore, the value of $\Delta G^o$ for the reaction is -959.1 kJ

3 0

3 years ago

What is the same for all of the drilling sites we examined?

yarga [219]

Answer:

the same is what is this question like what did u exame

4 0

3 years ago

Read 2 more answers

What is the first basic metal on the periodic table

kherson [118]

The first basic metals on the periodic table are alkali metals.

4 0

3 years ago

When maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas write the balanved equati

babunello [35]

Answer: When maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas then balanced equation is $MnO_{2} + 4HCl \rightarrow 2H_{2}O + MnCl_{2} + Cl_{2}$ .

Explanation:

The word equation is given as maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas.

Now, in terms of chemical formulae this reaction equation will be as follows.

$MnO_{2} + HCl \rightarrow H_{2}O + MnCl_{2} + Cl_{2}$

Here, number of atoms on reactant side are as follows.

Mn = 1
O = 2
H = 1
Cl = 1

Number of atoms on product side are as follows.

Mn = 1
O = 1
H = 2
Cl = 4

To balance this equation, multiply HCl by 4 on reactant side and multiply $H_{2}O$ by 2 on product side. Therefore, the equation can be rewritten as follows.

$MnO_{2} + 4HCl \rightarrow 2H_{2}O + MnCl_{2} + Cl_{2}$

Hence, number of atoms on reactant side are as follows.

Mn = 1
O = 2
H = 4
Cl = 4

Number of atoms on product side are as follows.

Mn = 1
O = 2
H = 4
Cl = 4

Since, this equation contains same number of atoms on both reactant and product side. Therefore, this equation is now balanced equation.

Thus, we can conclude that when maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas then balanced equation is $MnO_{2} + 4HCl \rightarrow 2H_{2}O + MnCl_{2} + Cl_{2}$ .

6 0

3 years ago

Platinum crystallizes in a face-centered cubic cell. the density of platinum is 21.4 g/cm3. calculate the radius of the platinum

IgorLugansk [536]

1 mole of platinum has a mass of 195 g therefore 1 atom will have a mass of 195 g /(6.02 ×10^23) = 3.239 × 10^-22 g
Density is given by dividing mass by volume, thus to get volume, mass is divided by density.
The volume = (3.239 × 10^-22)/21.4
= 1.514 × 10^-23 cm³
But volume of a sphere is given by 4/3πr³
Therefore, r³ = 3.6129 × 10^-24
r = ∛(3.6129 × 10^-24)
= 1.534 × 10^ -8 cm
Therefore, the radius of the platinum atom is 1.534 × 10^-8 cm

7 0

3 years ago