Question

Please solve this. Thermodynamics class 11..Please ╥﹏╥​

Answer 1

Three reactions are given.

$\small\text{ \longrightarrow\tt{C_{(s)}+O_{2(g)}\to CO_{2(g)};\quad\Delta H^o_{c(i)}=-393.5\ kJ\quad\dots(i)} }$

$\small\text{ \longrightarrow\tt{H_{2(g)}+\dfrac{1}{2}\,O_{2(g)}\to H_2O_{(l)};\quad\Delta H^0_{c(ii)}=-285.9\ kJ\quad\dots(ii)} }$

$\small\text{ \longrightarrow\tt{C_2H_{4(g)}+3O_{2(g)}\to2CO_{2(g)}+2H_2O_{(l)};\quad\Delta H^0_{c(iii)}=-1430.9\ kJ\quad\dots(iii)} }$

We need to calculate the enthalpy of change of,

$\small\text{ \longrightarrow\tt{2C_{(s)}+2H_{2(g)}\to C_2H_{4(g)};\quad\Delta H^0_{c(iv)}=\,?\quad\dots(iv)} }$

In this reaction 2 moles of $\tt{C_{(s)}}$ is used but one mole of $\tt{C_{(s)}}$ is used in (i). Also $\tt{C_{(s)}}$ is in same side (reactant side) in both the reactions. So (i) should be multiplied by 2.

In this reaction 2 moles of $\tt{H_{2(g)}}$ is used but one mole is used in (ii). Also $\tt{H_{2(g)}}$ is in same side (reactant side) in both the reactions. So (ii) should be multiplied by 2.

One mole of $\tt{C_2H_{4(g)}}$ is evolved in our reaction as well as in (iii), but the sides are different; $\tt{C_2H_{4(g)}}$ is in reactant side of (iii) but in product side of our reaction. So (iii) should be multiplied by -1.

Hence,

$\longrightarrow\tt{(iv)}=\tt{2(i)}+\tt{2(ii)}-\tt{(iii)}$

and so, the enthalpy of change of our reaction is given by,

$\longrightarrow\tt{\Delta H^0_{c(iv)}=2\Delta H^0_{c(i)}+2\Delta H^0_{c(ii)}-\Delta H^0_{c(iii)}}$

$\small\text{ \longrightarrow\tt{\Delta H^0_{c(iv)}=2(-393.5)+2(-285.9)-(-1430.9)} }$

$\longrightarrow\underline{\underline{\tt{\Delta H^0_{c(iv)}=72.1\ kJ}}}$