Problem 5
A graph of the logarithm of the equilibrium constant for a reaction versus \(\frac{1}{T}\) is linear but can have either a negative slope or a positive slope, depending on the reaction, as was observed here (Chapter 14). However, the graph of the logarithm of the rate constant for a reaction versus \(\frac{1}{T}\) has a neg ative slope for essentially every reaction. Using equilibrium arguments, explain why the graph for the rate constant must have a negative slope.
Problem 7
We found that the rate law for the reaction \(H_{2}(g)+I_{2}(g) \rightarrow 2 H I(g)\) is Rate \(=k\left[H_{2}\right]\left[I_{2}\right]\). Therefore, the reaction is second order overall but first order in \(H_{2}\). Imagine that we start with \(\left[H_{2}\right]_{0}=\left[I_{2}\right]_{0}\) and we measure \(\left[H_{2}\right]\) ver sus time. Will a graph of \(\ln \left(\left[H_{2}\right]\right)\) versus time be linear or will a graph of \(\frac{1}{\left[H_{2}\right.}\) versus time be linear? Explain your reasoning.
