QuestionJuly 24, 2025

2N_(2)O_(5)(aq)arrow 4NO_(2)(aq)+O_(2)(g) Which choice represents the relative rate expression between N_(2)O_(5) and NO_(2) ? 1 Rate=-(1)/(2)(Delta [N_(2)O_(5)])/(Delta t)=(1)/(4)(Delta [NO_(2)])/(Delta t) 2. Rate=(Delta [N_(2)O_(5)])/(Delta t)=(Delta [NO_(2)])/(Delta t) 3 Rate=-(1)/(4)(Delta [N_(2)O_(5)])/(Delta t)=(1)/(2)(Delta [NO_(2)])/(Delta t) 4 Rate=2(Delta [N_(2)O_(5)])/(Delta t)=4(Delta [NO_(2)])/(Delta t)

2N_(2)O_(5)(aq)arrow 4NO_(2)(aq)+O_(2)(g) Which choice represents the relative rate expression between N_(2)O_(5) and NO_(2) ? 1 Rate=-(1)/(2)(Delta [N_(2)O_(5)])/(Delta t)=(1)/(4)(Delta [NO_(2)])/(Delta t) 2. Rate=(Delta [N_(2)O_(5)])/(Delta t)=(Delta [NO_(2)])/(Delta t) 3 Rate=-(1)/(4)(Delta [N_(2)O_(5)])/(Delta t)=(1)/(2)(Delta [NO_(2)])/(Delta t) 4 Rate=2(Delta [N_(2)O_(5)])/(Delta t)=4(Delta [NO_(2)])/(Delta t)
2N_(2)O_(5)(aq)arrow 4NO_(2)(aq)+O_(2)(g) Which choice represents the relative rate expression between N_(2)O_(5) and NO_(2) ? 1 Rate=-(1)/(2)(Delta [N_(2)O_(5)])/(Delta t)=(1)/(4)(Delta [NO_(2)])/(Delta t) 2. Rate=(Delta [N_(2)O_(5)])/(Delta t)=(Delta [NO_(2)])/(Delta t) 3 Rate=-(1)/(4)(Delta [N_(2)O_(5)])/(Delta t)=(1)/(2)(Delta [NO_(2)])/(Delta t) 4 Rate=2(Delta [N_(2)O_(5)])/(Delta t)=4(Delta [NO_(2)])/(Delta t)

Solution
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Answer

1 Rate=-\frac {1}{2}\frac {\Delta [N_{2}O_{5}]}{\Delta t}=\frac {1}{4}\frac {\Delta [NO_{2}]}{\Delta t} Explanation 1. Identify the stoichiometric coefficients The reaction is 2N_{2}O_{5} \rightarrow 4NO_{2} + O_{2}, with coefficients 2 for N_{2}O_{5} and 4 for NO_{2}. 2. Write the rate expression The rate of disappearance of N_{2}O_{5} is -\frac{1}{2}\frac{\Delta [N_{2}O_{5}]}{\Delta t}, and the rate of formation of NO_{2} is \frac{1}{4}\frac{\Delta [NO_{2}]}{\Delta t}. 3. Relate the rates using stoichiometry According to stoichiometry, -\frac{1}{2}\frac{\Delta [N_{2}O_{5}]}{\Delta t} = \frac{1}{4}\frac{\Delta [NO_{2}]}{\Delta t}.

Explanation

1. Identify the stoichiometric coefficients<br /> The reaction is $2N_{2}O_{5} \rightarrow 4NO_{2} + O_{2}$, with coefficients 2 for $N_{2}O_{5}$ and 4 for $NO_{2}$.<br /><br />2. Write the rate expression<br /> The rate of disappearance of $N_{2}O_{5}$ is $-\frac{1}{2}\frac{\Delta [N_{2}O_{5}]}{\Delta t}$, and the rate of formation of $NO_{2}$ is $\frac{1}{4}\frac{\Delta [NO_{2}]}{\Delta t}$.<br /><br />3. Relate the rates using stoichiometry<br /> According to stoichiometry, $-\frac{1}{2}\frac{\Delta [N_{2}O_{5}]}{\Delta t} = \frac{1}{4}\frac{\Delta [NO_{2}]}{\Delta t}$.
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