Thermodynamics Test - 18

When beaten by a hammer the material heats up due to the impact. As the work done by this heat is zero there is a net increase in internal energy.

$$NH_3 + (3/2)CuO\longrightarrow \frac{1}{2}N_2(g)+ (3/2)H2O(l)+ (3/2)Cu(s) ; \:  \Delta H=?$$
                   $$\Delta H^{\circ}=\Sigma H^{\circ}_{Products}-\Sigma H^{\circ}_{Reactants}$$
                            $$\displaystyle =[\frac{1}{2}\times H^{\circ}_{N_2}+( 3 / 2)\times H^{\circ}_{H_2O(l)}+~(3 /2) H^{\circ}_{Cu}]-[(3/2)H^{\circ}_{CuO}+H^{\circ}_{NH_3}]$$
                            $$=[0+ (3/2)\times (-285)+ (3/2)\times 0]-[(3/2) \times(-155)+ (-46)]$$
                            $$=-149\,kJ$$
                     $$[\because H^{\circ} for \: pure \: element = 0 \: and \: H^{\circ}_{compound} =H^{\circ}_{formation}] $$
$$\because 17 \: g \: NH_3 \: brings \: in \: change \: in \: \Delta H = 149 \: kJ$$
$$\because 6.8 \: g \: NH_3 \: brings \: in \: change \: in \: \Delta H = (149 \times 6.8)/17 = 59.6 \,kJ$$
$$\therefore  \Delta H for \: 6.89 \: g \: NH_3 = - 59.6 \: kJ$$

Bond energy $$={ BE }_{ { X }_{ e }-F }=34kCal/mol$$
So, $${ X }_{ e }{ F }_{ 4 }=4 ({ X }_{ e }-F)$$ bonds
$$\Rightarrow$$ total bond energy$$=4\times 34kCal/mol \Rightarrow{ BE }_{ { X }_{ e }-{ F }_{ 4 } }=136kCal/mol$$
Now this amount of bond energy is liberated during formation of $${ X }_{ e }{ F }_{ 4 }$$   $$\Delta _{ f }{ H }_{ { X }_{ e }{ F }_{ 4 } }=-136kCal/mol$$
$${ X }_{ e }+4F\rightarrow { X }_{ e }{ F }_{ 4 }$$   $$\Delta _{ f }{ H }_{ { X }_{ e }{ F }_{ 4 } }=-136kCal/mol$$   ...(1)
Now $$I{ E }_{ 1 }$$ of $${ X }_{ e }=279kCal/mol$$
So $${ X }_{ e }\rightarrow { { X }_{ { e } } }^{ + }+{ e }^{ - }$$   $$\Delta H=+I{ E }_{ 1 }$$    ...(2)
Now, $$EA$$ of $$F=85kCal/mol$$
$$F\rightarrow { F }^{ - }-{ e }^{ - }$$   $$\Delta H=-{ E }A$$     ...(3)
(energy is released in this process)
and $$2F\rightarrow{ F }_{ 2 }$$   $$\Delta K=-{ e }_{ F-F }$$   ...(4)
Now $$(1)$$ becomes$$\Rightarrow { X }_{ e }{ F }_{ 4 }\rightarrow { X }_{ e }+4F$$   $$\Delta _{ f }{ H }_{ { X }_{ e }{ F }_{ 4 } }=136kCal/mol$$    ...(5)
Now, $$(2)+(3)+(4)+(5)$$ gives $${ X }_{ e }{ F }_{ 4 }+{ X }_{ e }+3F\rightarrow { X }_{ e }+4F+{ X }_{ e }^{ + }+{ e }^{ - }+{ F }^{ - }-{ e }^{ - }+{ F }_{ 2 }$$
$$\Delta H=\Delta _{ f }{ H }_{ { X }_{ e }{ F }_{ 4 } }+\left( H{ E }_{ 1 } \right) +\left( -EA \right) +\left( -{ e }_{ F-F } \right) =136+279+(-85)+(-38)=292$$
$$\Rightarrow { X }_{ e }{ F }_{ 4 }={ X }_{ e }^{ + }+{ F }^{ - }+{ F }_{ 2 }+F$$
$$\Delta H=292kCal/mol$$

Heat capacity of gases increase at room temperature by increasing the atomicity.

The heat required to raise the temperature of a body by 1 K is called thermal capacity.
In other words, when q is the heat supplied to the body and the temperature raises by 1 K, then the thermal capacity of body is q.

  $$C_D\longrightarrow C_G; \:               \Delta H= - 1.89$$ ...( i)
   $$C_D+ O_2\longrightarrow CO_2; \:          \Delta H= - q_1$$    ...( ii)
   $$C_G+ O_2\longrightarrow CO_2; \:          \Delta H= - q_2$$   ...( iii)
Subtract equation (iii) from equation (ii).
$$C_D\longrightarrow C_G; \:               \Delta H= - q_1+ q_2$$...(iv)
From equations (i) and (iv)
$$q_1+ q_2=-1.89$$
or   $$q_2-q_1=-1.89 \: for \: 12 \: g \: C_{D\rightarrow G} $$
Thus, $$for \: 6 \: g \: C_{D\rightarrow G} $$
     $$\displaystyle q_2-q_1=\frac{-1.89}{2}= -0 .945 \:kJ$$

For the reaction,
      $$C+ O_2\longrightarrow CO_2; \:      \Delta H=-393\:J$$ ......(1)
 $$2Zn + O_2\longrightarrow 2ZnO;\:    \Delta H=-412J\:J$$ .......(2)
Reverse reaction (2) and add it to reaction (1)
$$C + 2ZnO\longrightarrow CO_2+2Zn; \:        \Delta H =+19.0$$
Hence, Zn liberates more heat than carbon during oxidation

Hess's law is related to change in heat during a reaction.
It states that the total change in heat enthalpy during the complete course of reaction is same whether the change is brought in one step or in several steps by one method or other method.

The decrease in Gibbs energy results in useful work done by the system.