Answer:
(a) -0.00017 M/s;
(b) 0.00034 M/s
Explanation:
(a) Rate of a reaction is defined as change in molarity in a unit time, that is:
[tex]r = \frac{\Delta c}{\Delta t}[/tex]
Given the following reaction:
[tex]2 N_2O_5 (g)\rightleftharpoons 4 NO_2 (g) + O_2 (g)[/tex]
We may write the rate expression in terms of reactants firstly. Since reactants are decreasing in molarity, we're adding a negative sign. Similarly, if we wish to look at the overall reaction rate, we need to divide by stoichiometric coefficients:
[tex]r = -\frac{\Delta [N_2O_5]}{2 \Delta t}[/tex]
Reaction rate is also equal to the rate of formation of products divided by their coefficients:
[tex]r = \frac{\Delta [NO_2]}{4 \Delta t} = \frac{\Delta [O_2]}{\Delta t}[/tex]
Let's find the rate of disappearance of the reactant firstly. This would be found dividing the change in molarity by the change in time:
[tex]r_{N_2O_5} = \frac{0.066 M - 0.100 M}{200.00 s - 0.00 s} = -0.00017 M/s[/tex]
(b) Using the relationship derived previously, we know that:
[tex]-\frac{\Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{4 \Delta t}[/tex]
Rate of appearance of nitrogen dioxide is given by:
[tex]r_{NO_2} = \frac{\Delta [NO_2]}{\Delta t}[/tex]
Which is obtained from the equation:
[tex]-\frac{\Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{4 \Delta t}[/tex]
If we multiply both sides by 4, that is:
[tex]-\frac{4 \Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{\Delta t}[/tex]
This yields:
[tex]r_{NO_2} = \frac{\Delta [NO_2]}{\Delta t} = -2\frac{\Delta [N_2O_5]}{ \Delta t} = -2\cdot (-0.00017 M/s) = 0.00034 M/s[tex]
