Ejercicio 1

$$\begin{aligned} T_K &= {T_C} + {273.1} \\ &= 200.0000 \ \mathrm{°C} + 273.1500 \ \mathrm{} \\ &= 473.1500 \ \mathrm{K} \\ \end{aligned}$$

$$\begin{aligned} n_\text{freón} &= \frac{ {P} \cdot {V} }{ {R} \cdot {T_K} } \\ &= \frac{ 1.0395 \ \mathrm{atm} \cdot 8.2900 \ \mathrm{L} }{ 0.0821 \ \mathrm{atm\ L / mol\ K} \cdot 473.1500 \ \mathrm{K} } \\ &= 0.2219 \ \mathrm{mol} \\ \end{aligned}$$

$$\begin{aligned} \text{masa molar freón} &= \frac{ {g_{\text{freón}}} }{ {n_\text{freón}} } \\ &= \frac{ 26.8000 \ \mathrm{g} }{ 0.2219 \ \mathrm{mol} } \\ &= 120.7484 \ \mathrm{g/mol} \\ \end{aligned}$$

ejercicio 2

$$\begin{aligned} V_{L} &= {V_{m^{3}}} \cdot {1000} \\ &= 0.4000 \ \mathrm{m^{3}} \cdot 1000.0000 \ \mathrm{} \\ &= 400.0000 \ \mathrm{L} \\ \end{aligned}$$

$$\begin{aligned} T_K &= {T_c} + {273.1} \\ &= 520.0000 \ \mathrm{°C} + 273.1500 \ \mathrm{} \\ &= 793.1500 \ \mathrm{K} \\ \end{aligned}$$

$$\begin{aligned} \text{mol}_{CH_4} &= {g_{CH_4}} \cdot \frac{ {1 mol} }{ {g CH4} } \\ &= { 4.585 \times 10^{4} } \ \mathrm{g} \cdot \frac{ 1.0000 \ \mathrm{mol} }{ 16.0430 \ \mathrm{g} } \\ &= 2857.9443 \ \mathrm{mol} \\ \end{aligned}$$

$$\begin{aligned} P_{\text{ideal}} &= \frac{ {\text{mol}_{CH_4}} \cdot {R} \cdot {T_K} }{ {V_{L}} } \\ &= \frac{ 2857.9443 \ \mathrm{mol} \cdot 0.0821 \ \mathrm{atm\ L / mol\ K} \cdot 793.1500 \ \mathrm{K} }{ 400.0000 \ \mathrm{L} } \\ &= 465.0126 \ \mathrm{atm} \\ \end{aligned}$$

$$\begin{aligned} P_{\text{real}} &= \frac{ {\text{mol}_{CH_4}} \cdot {R} \cdot {T_K} }{ {V_{L}} - {\text{mol}_{CH_4}} \cdot {b} } - \frac{ {\left( {\text{mol}_{CH_4}} \right)}^{ {2} } \cdot {a} }{ {\left( {V_{L}} \right)}^{ {2} } } \\ &= \frac{ 2857.9443 \ \mathrm{mol} \cdot 0.0821 \ \mathrm{atm\ L / mol\ K} \cdot 793.1500 \ \mathrm{K} }{ 400.0000 \ \mathrm{L} - 2857.9443 \ \mathrm{mol} \cdot 0.0428 \ \mathrm{L/\text{mol}} } - \frac{ {\left( 2857.9443 \ \mathrm{mol} \right)}^{ 2.0000 \ \mathrm{} } \cdot 2.2830 \ \mathrm{L^2 bar/\text{mol}^2} }{ {\left( 400.0000 \ \mathrm{L} \right)}^{ 2.0000 \ \mathrm{} } } \\ &= 553.1712 \ \mathrm{atm} \\ \end{aligned}$$

ejercicio 3

$$\rm Zn + 2HCl\to ZnCl_2 + H_2$$

$$\begin{aligned} V_L &= {V_{mL}} \cdot \frac{ {1\ L} }{ {1000\ mL} } \\ &= 90.0000 \ \mathrm{mL} \cdot \frac{ 1.0000 \ \mathrm{L} }{ 1000.0000 \ \mathrm{mL} } \\ &= 0.0900 \ \mathrm{L} \\ \end{aligned}$$

$$\begin{aligned} P_{atm} &= {P_{mmHg}} \cdot \frac{ {1\ atm} }{ {760\ mmHg} } \\ &= 560.0000 \ \mathrm{mmHg} \cdot \frac{ 1.0000 \ \mathrm{atm} }{ 760.0000 \ \mathrm{mmHg} } \\ &= 0.7368 \ \mathrm{atm} \\ \end{aligned}$$

$$\begin{aligned} T_K &= {273.1} + {T_C} \\ &= 273.1500 \ \mathrm{} + 22.0000 \ \mathrm{°C} \\ &= 295.1500 \ \mathrm{K} \\ \end{aligned}$$

$$\begin{aligned} mol\ H_2 &= \frac{ {P_{atm}} \cdot {V_L} }{ {R} \cdot {T_K} } \\ &= \frac{ 0.7368 \ \mathrm{atm} \cdot 0.0900 \ \mathrm{L} }{ 0.0821 \ \mathrm{atm\ L / mol\ K} \cdot 295.1500 \ \mathrm{K} } \\ &= 0.0027 \ \mathrm{mol\ H_2} \\ \end{aligned}$$

$$\begin{aligned} g\ Zn &= {mol\ H_2} \cdot \frac{ {1\ mol\ Zn} }{ {1\ mol\ H_2} } \cdot \frac{ {65.38\ g\ Zn} }{ {1\ mol\ Zn} } \\ &= 0.0027 \ \mathrm{mol\ H_2} \cdot \frac{ 1.0000 \ \mathrm{mol\ Zn} }{ 1.0000 \ \mathrm{mol\ H_2} } \cdot \frac{ 65.3800 \ \mathrm{g\ Zn} }{ 1.0000 \ \mathrm{mol\ Zn} } \\ &= 0.1790 \ \mathrm{g\ Zn} \\ \end{aligned}$$

$$\begin{aligned} g\ impuros\ Zn &= {g\ Zn} \cdot \frac{ {100\ g\ impuros\ Zn} }{ {68\ g\ Zn} } \\ &= 0.1790 \ \mathrm{g\ Zn} \cdot \frac{ 100.0000 \ \mathrm{g\ impuros\ Zn} }{ 68.0000 \ \mathrm{g\ Zn} } \\ &= 0.2633 \ \mathrm{g\ impuros\ Zn} \\ \end{aligned}$$

ejercicio 4

$$\begin{aligned} g\ PH_3 &= {mg\ PH_3} \cdot \frac{ {1\ g} }{ {1000\ mg} } \\ &= { 7.0 \times 10^{-5} } \ \mathrm{mg} \cdot \frac{ 1.0000 \ \mathrm{g} }{ 1000.0000 \ \mathrm{mg} } \\ &= { 7.0 \times 10^{-8} } \ \mathrm{g\ PH_3} \\ \end{aligned}$$

$$\begin{aligned} n_{PH_3} &= {g\ PH_3} \cdot \frac{ {1\ mol\ PH_3} }{ {33.998\ g\ PH_3} } \\ &= { 7.0 \times 10^{-8} } \ \mathrm{g\ PH_3} \cdot \frac{ 1.0000 \ \mathrm{mol\ PH_3} }{ 33.9980 \ \mathrm{g\ PH_3} } \\ &= { 2.0589 \times 10^{-9} } \ \mathrm{n_{PH_3}} \\ \end{aligned}$$

$$\begin{aligned} T_K &= {T_C} + {273.1} \\ &= 25.0000 \ \mathrm{°C} + 273.1500 \ \mathrm{} \\ &= 298.1500 \ \mathrm{K} \\ \end{aligned}$$

$$\begin{aligned} P &= \frac{ {n_{PH_3}} \cdot {R} \cdot {T_K} }{ {V_L} } \\ &= \frac{ { 2.0589 \times 10^{-9} } \ \mathrm{n_{PH_3}} \cdot 0.0821 \ \mathrm{atm\ L / mol\ K} \cdot 298.1500 \ \mathrm{K} }{ 1.0000 \ \mathrm{L} } \\ &= { 5.0373 \times 10^{-8} } \ \mathrm{atm} \\ \end{aligned}$$

ejercicio 5

$$\begin{aligned} g_{C_2H_5Cl} &= {kg} \cdot \frac{ {1000\ g} }{ {1\ kg} } \\ &= 18.5000 \ \mathrm{kg} \cdot \frac{ 1000.0000 \ \mathrm{g} }{ 1.0000 \ \mathrm{kg} } \\ &= { 1.85 \times 10^{4} } \ \mathrm{g} \\ \end{aligned}$$

$$\begin{aligned} moles &= {g_{C_2H_5Cl}} \cdot \frac{ {1\ mol} }{ {64.512\ g} } \\ &= { 1.85 \times 10^{4} } \ \mathrm{g} \cdot \frac{ 1.0000 \ \mathrm{mol} }{ 64.5120 \ \mathrm{g} } \\ &= 286.7684 \ \mathrm{mol} \\ \end{aligned}$$

$$\begin{aligned} V_L &= {V_{m^3}} \cdot \frac{ {1000\ L} }{ {1\ m^3} } \\ &= 2.1000 \ \mathrm{m^3} \cdot \frac{ 1000.0000 \ \mathrm{L} }{ 1.0000 \ \mathrm{m^3} } \\ &= 2100.0000 \ \mathrm{L} \\ \end{aligned}$$

$$\begin{aligned} T_K &= {T_C} + {273.1} \\ &= 200.0000 \ \mathrm{°C} + 273.1500 \ \mathrm{} \\ &= 473.1500 \ \mathrm{K} \\ \end{aligned}$$

$$\begin{aligned} P_{\text{real}} &= \frac{ {moles} \cdot {R} \cdot {T_K} }{ {V_L} - {moles} \cdot {b} } - \frac{ {\left( {moles} \right)}^{ {2} } \cdot {a} }{ {\left( {V_L} \right)}^{ {2} } } \\ &= \frac{ 286.7684 \ \mathrm{mol} \cdot 0.0821 \ \mathrm{atm\ L / mol\ K} \cdot 473.1500 \ \mathrm{K} }{ 2100.0000 \ \mathrm{L} - 286.7684 \ \mathrm{mol} \cdot 0.0865 \ \mathrm{L/\text{mol}} } - \frac{ {\left( 286.7684 \ \mathrm{mol} \right)}^{ 2.0000 \ \mathrm{} } \cdot 11.0500 \ \mathrm{L^2 bar/\text{mol}^2} }{ {\left( 2100.0000 \ \mathrm{L} \right)}^{ 2.0000 \ \mathrm{} } } \\ &= 5.1592 \ \mathrm{atm} \\ \end{aligned}$$

ejercicio 7 - nitrato de amonio

$$\begin{aligned} n_{NH_4NO_3} &= {g} \cdot \frac{ {1\ mol\ NH_4NO_3} }{ {80.043\ g\ NH_4NO_3} } \\ &= 3.8800 \ \mathrm{g} \cdot \frac{ 1.0000 \ \mathrm{mol\ NH_4NO_3} }{ 80.0430 \ \mathrm{g\ NH_4NO_3} } \\ &= 0.0485 \ \mathrm{mol\ NH_4NO_3} \\ \end{aligned}$$

$$\begin{aligned} \Delta T &= {T_{fin}} - {T_{ini}} \\ &= 18.4000 \ \mathrm{°C} - 23.0000 \ \mathrm{°C} \\ &= \left( -4.6000 \right) \ \mathrm{°C} \\ \end{aligned}$$

$$\begin{aligned} q_{H_2O} &= {g_{H_2O}} \cdot {CE_{H_2O}} \cdot {\Delta T} \\ &= 60.0000 \ \mathrm{g} \cdot 75.3850 \ \mathrm{J/mol\ °C} \cdot \left( -4.6000 \right) \ \mathrm{°C} \\ &= \left( -2.0806 \times 10^{4} \right) \ \mathrm{J} \\ \end{aligned}$$

$$\begin{aligned} q_{disol} &= \left( - {q_{H_2O}} \right) \\ &= \left( - \left( -2.0806 \times 10^{4} \right) \ \mathrm{J} \right) \\ &= { 2.0806 \times 10^{4} } \ \mathrm{J} \\ \end{aligned}$$

$$\begin{aligned} \Delta H_{disol} &= \frac{ {q_{disol}} }{ {n_{NH_4NO_3}} } \cdot \frac{ {1\ kJ/mol} }{ {1000\ J/mol} } \\ &= \frac{ { 2.0806 \times 10^{4} } \ \mathrm{J} }{ 0.0485 \ \mathrm{mol\ NH_4NO_3} } \cdot \frac{ 1.0000 \ \mathrm{kJ/mol} }{ 1000.0000 \ \mathrm{J/mol} } \\ &= 429.2256 \ \mathrm{kJ/mol} \\ \end{aligned}$$

ejercicio 8 - cobre

$$\begin{aligned} \Delta T &= {T_{fin}} - {T_{ini}} \\ &= 88.5000 \ \mathrm{°C} - 25.0000 \ \mathrm{°C} \\ &= 63.5000 \ \mathrm{°C} \\ \end{aligned}$$

$$\begin{aligned} q_{Cu} &= {g_{Cu}} \cdot {CE_{Cu}} \cdot {\Delta T} \\ &= 1420.0000 \ \mathrm{g} \cdot 0.3850 \ \mathrm{J/g °C} \cdot 63.5000 \ \mathrm{°C} \\ &= { 3.4715 \times 10^{4} } \ \mathrm{J} \\ \end{aligned}$$

$$\begin{aligned} q_{Cu} &= {q_{Cu}} \cdot \frac{ {1\ kJ} }{ {1000\ J} } \\ &= { 3.4715 \times 10^{4} } \ \mathrm{J} \cdot \frac{ 1.0000 \ \mathrm{kJ} }{ 1000.0000 \ \mathrm{J} } \\ &= 34.7155 \ \mathrm{kJ} \\ \end{aligned}$$

Ejercicio 10 - ley de Hess

$$ \begin{array}{rclcl} N_2O & \to & N_2 + \frac{1}{2} O_2 & & \frac{-163.2}{2} kJ\\[1em] NO_2 & \to & NO + \frac{1}{2} O_2 & & \frac{113.1}{2} kJ\\[1em] N_2 + O_2 & \to & 2NO && 180.7 kJ\\[1em] N_2O + NO_2 & \to & 3NO && \frac{-163.2+113.1}{2} + 180.7 = 155.65 kJ \end{array} $$

Ejercicio 11 - entropía

$$ \rm 2NO + O_2 \to 2NO_2 $$

se tendrá mayor entropía en los reactivos, porque son dos partículas separadas ($NO$ y $O_2$), mientras que en productos hay sólo una partícula ($NO_2$)