Question

Zinc is commonly found in nature in the form of the mineral sphalerite (ZnS). A step in the smelting of zinc is the rusting of sphalerite with oxygen to produce zinc oxide

$2ZnS\left(s\right)+3O_2\left(g\right)\to 2ZnO\left(s\right)+2S{O}_2\left(g\right)$

(a) Calculate the standard enthalpy change$∆H^{\circ }$for this reaction, using data from Appendix D.

(b) Calculate the heat absorbed when 3.00 metrics tons ( 1 metric ton =10³kg)of sphalerite is roasted under constant-pressure conditions.

Short answer

(a) The enthalpy change value $∆H^{\circ }=-878.26kJ$.

(b) The amount of heat evolved q_p = -1.35$\times {10}^{7}kJ$.

Step-by-step solution

Given data

$2\text{ZnS}\left(s\right)+3{\text{O}}_2\left(g\right)\to 2\text{ZnO}\left(s\right)+2{\text{SO}}_2\left(g\right)$

$\Delta H_{\text{f}}^e$Value of $\text{ZnO}$is $-348.28{\text{kJmol}}^{-1}$.

$\Delta H_{\text{f}}^e$Value of O₂ is 0.

$\Delta H_{\text{f}}^e$Value of $\text{ZnS}$Is $-205.98{\text{kJmol}}^{-1}$.

$\Delta H_{\text{f}}^e$Value of ${\text{SO}}_2$is $-296.83{\text{kJmolm}}^{-1}$.

$\Delta H_{\text{f}}^e$Represents the standard enthalpy value.

(b) localid="1663769733762" $3.00\mathrm{metric}\mathrm{tons}$width="153">(1metricton=103kg)

Concept of Enthalpy

Enthalpy change is amount of heat absorbed or evolved in a reaction when carried out at a constant pressure. It is generally calculated as the difference in energy required for bond breaking in a chemical reaction and energy gained in forming new bonds in a chemical reaction.

It is represented by$∆H$ .

Negative enthalpy change represents the exothermic reaction when energy is released from the reaction and positive enthalpy change represents an endothermic reaction in which energy is taken in from surroundings.

Calculate the enthalpy change ∆H∘ 

(a)

$Theenthalpychange∆H=n∆H_f^{\circ }\left\{Products\right)-n∆H_f^{\circ }\left\{Reac\tan ts\right)$

$2\text{ZnS}\left(s\right)+3{\text{O}}_2\left(g\right)\to 2\text{ZnO}\left(s\right)+2\text{SO}2\left(g\right)$

$\Delta H_{\text{f}}^e$value of $\text{ZnO}$is $-348.28{\text{kJmol}}^{-1}$.

$\Delta H_{\text{f}}^e$ Value of O₂ is 0.

$\Delta H_{\text{f}}^e$Value of ZnS is $-205.98{\text{kJmol}}^{-1}$.

$\Delta H_{\text{f}}^e$ Value of SO₂is $-296.83{\text{kJmolm}}^{-1}$.

$\Delta H_{\text{f}}^e$Represents the standard enthalpy value.

role="math" localid="1663770975183" $$\begin{gathered} ∆H^{\circ }=\left(2mol\right)∆H_f^{\circ }\left[ZnO\right(s\left)\right]+\left(2mol\right)∆H_f^{\circ }\left[S{O}_2\right(g\left)\right]-\left\{\right(2mol)∆H_f^{\circ }[ZnS\left(s\right)]+(3mol)∆H_f^{\circ }[O_2\left(g\right)\left]\right\} \\ =\left(2mol\right)(-348.28kJmo{l}^{-1})+\left(2mol\right)(-296.83kJmo{l}^{-1})-\left(2mol\right)(-205.98kJmo{l}^{-1})+\left(3mol\right)\times 0 \\ =-696.56kJ-593.66kJ+411.96kJ \\ =-878.26kJ \end{gathered}$$

Calculate the molar mass of ZnS

(b)

Solar mass:

$$\begin{gathered} \text{Molar mass of ZnS}=\left(\text{atomic mass of Zn}\right)+(\text{atomic mass of}\text{S}) \\ M\left(\text{ZnS}\right)=65.38{\text{gmol}}^{-1}+32.06{\text{gmol}}^{-1} \\ =97.44{\text{gmol}}^{-1} \end{gathered}$$

Calculate the moles of ZnS 

Calculate the moles of based on Molar mass of (ZnS).

$$\begin{gathered} Molesof\text{ZnS}=\frac{1.00\text{gZnS}}{97.44{\text{gmol}}^{-1}} \\ =3.079\times {10}^4\text{molZnS} \end{gathered}$$

Calculate the value of qp.

$∆H^{\circ }=-878.26kJ$is associated with 2 moles of ZnS, so heat evolved associated with $3.079\times {10}^4\text{mol}\text{ZnS}$is:

$$\begin{gathered} \Delta H=3.079\times {10}^4\text{molZnS}\times \frac{-878.26\text{kJ}}{2\text{moles of ZnS}} \\ =-1.35\times {10}^7\text{kJ} \end{gathered}$$

At constant pressure the heat absorbed $q_{\text{p}}=\Delta H$.

Amount of heat evolved $q_p=-1.35\times {10}^7\text{k}J$.