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Chapter 13: Problem 35

Compound A converts to compound \(\mathrm{B} ; \Delta E_{\mathrm{rxn}}\) is \(-100 \mathrm{~kJ} / \mathrm{mol}\). Is compound \(\mathrm{B}\) at a higher or lower energy level than compound A? By how much?

Short Answer

Expert verified
Compound B is at a lower energy level than compound A, and the energy difference between them is 100 kJ/mol.

Step by step solution

01

Understand the meaning of the reaction energy value

The reaction energy value given is negative, which means that energy is being released during the reaction. Specifically, it states that the reaction energy \(\Delta E_{\mathrm{rxn}}\) is \(-100 \mathrm{~kJ/mol}\).
02

Identify whether compound B is at a higher or lower energy level than compound A

Since the reaction energy is negative, it indicates that energy is released when compound A converts to compound B. This implies that compound B is at a lower energy level than compound A.
03

Calculate the energy difference between compound A and compound B

The energy difference between compounds A and B is given by the reaction energy. In this case, \(\Delta E_{\mathrm{rxn}} = -100 \mathrm{~kJ/mol}\). Therefore, compound B is lower in energy than compound A by 100 kJ/mol. So, the answer is that compound B is at a lower energy level than compound A, and the energy difference is 100 kJ/mol.

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Most popular questions from this chapter

What is meant by the mechanism of a chemical reaction?

A reaction \(\mathrm{A}+\mathrm{B} \rightarrow\) Product is run in a balloon. (Both A and B are gases.) The balloon has a volume of 1 L and is initially loaded with 1 mole of \(\mathrm{A}\) and \(1 \mathrm{~mole}\) of \(\mathrm{B}\). The reaction has the rate law Rate \(=k[\mathrm{~A}]\) The reaction is run again using the same amount of reactants, but this time in a balloon that has a volume of \(0.5 \mathrm{~L}\). How much faster will the reaction proceed in the smaller balloon? Explain your answer.

Given the rate data below from a series of kinetics experiments, determine the orders for the following reaction, and state the overall order of the reaction: \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)+3 \mathrm{I}^{-}(a q)+2 \mathrm{H}^{+}(a q) \rightarrow \mathrm{I}_{3}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\) $$\begin{array}{cccc} \text { Experiment }\left[\mathbf{H}_{2} \mathrm{O}_{2}\right] & {\left[\mathbf{I}^{-}\right]} & {\left[\mathbf{H}^{+}\right]} & \text {Rate }(\mathbf{M} / \mathbf{s}) \\ \hline 1 & 0.010 \mathrm{M} & 0.010 \mathrm{M} & 0.00050 \mathrm{M} & 1.15 \times 10^{-6} \\ 2 & 0.020 \mathrm{M} & 0.010 \mathrm{M} & 0.00050 \mathrm{M} & 2.30 \times 10^{-6} \\ 3 & 0.010 \mathrm{M} & 0.020 \mathrm{M} & 0.00050 \mathrm{M} & 2.30 \times 10^{-6} \\ 4 & 0.010 \mathrm{M} & 0.010 \mathrm{M} & 0.00100 \mathrm{M} & 1.15 \times 10^{-6} \end{array}$$

If a reaction rate has a second-order dependence on a given reactant concentration, what will happen to the rate when the concentration of that reactant is doubled?

Using reaction-energy profiles, plot two exothermic reactions that have the same \(\Delta E_{\mathrm{rxn}}\), but make one reaction substantially faster than the other. Label the plots "fast" and "slow," and explain why you labeled them as you did.
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