The values of $$\Delta H$$ and $$\Delta S$$ for the reaction, $${C_{\left( {{\text{graphite}}} \right)}} + C{O_2}\left( g \right) \to 2CO\left( g \right)$$ are $$170\,kJ$$ and $$170\,J{K^{ - 1}},$$ respectively. This reaction will be spontaneous at
A.
710$$\,K$$
B.
910$$\,K$$
C.
1110$$\,K$$
D.
510$$\,K$$
Answer :
1110$$\,K$$
Solution :
$$\eqalign{
& {\text{Given,}}\,\Delta H = 170\,kJ = 170 \times {10^3}J \cr
& \Delta S = 170\,J{K^{ - 1}};\,T = ? \cr
& \Delta G = \Delta H - T\Delta S \cr
& {\text{For spontaneous reaction,}} \cr} $$
$$\Delta G < 0 \Rightarrow 0 < 170 \times {10^3} - T \times 170\,;$$ $$T > 1000$$
$$\therefore \,\,T = 1110\,K$$
Releted MCQ Question on Physical Chemistry >> Chemical Thermodynamics
Releted Question 1
The difference between heats of reaction at constant pressure and constant volume for the reaction : $$2{C_6}{H_6}\left( l \right) + 15{O_{2\left( g \right)}} \to $$ $$12C{O_2}\left( g \right) + 6{H_2}O\left( l \right)$$ at $${25^ \circ }C$$ in $$kJ$$ is
$${\text{The}}\,\Delta H_f^0\,{\text{for}}\,C{O_2}\left( g \right),\,CO\left( g \right)\,$$ and $${H_2}O\left( g \right)$$ are $$-393.5,$$ $$-110.5$$ and $$ - 241.8\,kJ\,mo{l^{ - 1}}$$ respectively. The standard enthalpy change ( in $$kJ$$ ) for the reaction $$C{O_2}\left( g \right) + {H_2}\left( g \right) \to CO\left( g \right) + {H_2}O\left( g \right)\,{\text{is}}$$