For the reaction $H_{2} + I_{2} \rightarrow 2 HI$,provide the potential energy profile diagram and prove that it is an exothermic reaction.
(N/A) The potential energy profile diagram shows the energy change during the reaction.
From the graph,it is clear that the potential energy of the products $(E_{P})$ is less than the potential energy of the reactants $(E_{R})$.
Since $H_{P} < H_{R}$,the change in enthalpy is given by:
$\Delta H = (H_{P} - H_{R}) < 0$ (negative value).
Because energy is released during the process,the reaction is exothermic.
From the graph,it is clear that the potential energy of the products $(E_{P})$ is less than the potential energy of the reactants $(E_{R})$.
Since $H_{P} < H_{R}$,the change in enthalpy is given by:
$\Delta H = (H_{P} - H_{R}) < 0$ (negative value).
Because energy is released during the process,the reaction is exothermic.
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$(2)$ $\frac{1}{8} S_{8(s)} + O_{2(g)} \rightarrow SO_{2(g)}$
$(3)$ $2 H_{2(g)} + O_{2(g)} \rightarrow 2 H_2O_{(l)}$
$(4)$ $2 C_{(g)} + 3 H_{2(g)} \rightarrow C_2H_{6(g)}$
MediumIIT 2019
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