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Listed in the table are forward and reverse rate constants for the reaction $$\ce{2NO (g) <=> N_2 (g) + O_2 (g)}$$ \begin{array}{ccc} \text{Temperature}/\pu{K} & k_f/\pu{M^-1s^-1} & k_r /\pu{M^-1s^-1} \\\hline 1400 & 0.29 & 1.1\times10^{-6}\\ 1500 & 1.3 & 1.4\times10^{-5}\\ \end{array} Is the reaction endothermic or exothermic?

In my book the solution is given as

$k_r$ increase more than $k_f$, this means that $E_a$ (reverse) is greater than $E_a$ (forward). The reaction is exothermic when $E_a$ (reverse) > $E_a$ (forward).

I can not understand what they mean by the given answer. Can anybody explain it in more detail?

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closed as off-topic by Zhe, Klaus-Dieter Warzecha, Todd Minehardt, Wildcat, ron Dec 29 '16 at 19:36

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  • $\begingroup$ My hint is use the van't Hoff equation. $\endgroup$ – Zhe Dec 29 '16 at 14:12
  • $\begingroup$ @Zhe I am not understanding how to use that $\endgroup$ – user123733 Dec 29 '16 at 16:00
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    $\begingroup$ The van't Hoff equation tells you what direction the equilibrium will shift as you change the temperature. It depends on whether the reaction is endothermic or exothermic. For endothermic reactions, as you increase temperature, the equilibrium shifts more towards products. The opposite is true for exothermic reactions. $\endgroup$ – Zhe Dec 29 '16 at 16:22
  • $\begingroup$ So @Zhe can we write -$\Delta$H =E$_a$(forward) - E$_a$(reverse) $\endgroup$ – user123733 Dec 29 '16 at 17:04
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    $\begingroup$ i.stack.imgur.com/GiJ0M.jpg $\endgroup$ – orthocresol Dec 29 '16 at 18:09