Question

Question:For the decomposition of gaseous dinitrogen pentaoxide,

$2{\mathrm{N}}_{5\left(\mathrm{g}\right)}\to 4{\mathrm{NO}}_{2\left(\mathrm{g}\right)}+{\mathrm{O}}_{2\left(\mathrm{g}\right),}$the rate constant is k $=2.8\times {10}^{-3}{\mathrm{s}}^{-1}\mathrm{at}60°\mathrm{C}.\mathrm{The}$initial concentration of ${\mathrm{N}}_2{\mathrm{O}}_5$is 1.58 mol/L.

(a) What $\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]$after 5.00 min?

(b) What fraction of the${\mathrm{N}}_2{\mathrm{O}}_5$has decomposed after 5.00 min?

Short answer

$\left(\mathrm{a}\right)$After 5.00 min, the concentration is equal to 0.68 mol/L.

$\left(\mathrm{b}\right)$After5.00 min, the fraction of decomposed is equal to 0.57.

Step-by-step solution

1:What is N2O5 after 5.00 min?

The rate constant, k, is expressed in units of,${\mathrm{s}}^{-1}$ implying a first-order reaction Consider$\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]$as the starting concentration and $\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]$as the concentration at time t. For a first-order process, use the integral rate law to determine the reactant concentration: