For the reaction $\mathrm{2\ N_2O_5(g) \longrightarrow 4\ NO_2(g) + O_2(g)}$ in liquid bromine, N$_2$O$_5$ disappears at a rate of 0.02 moles dm$^{-3}$ sec$^{-1}$. At what rate NO$_2$ and O$_2$ are formed? What would be the rate of reaction?

Question

Accepted Answer

> [!TLDR]
> N$_2$O$_5$ disappears at 0.02 mol dm$^{-3}$ s$^{-1}$; dividing by its coefficient 2 gives rate = 0.01; O$_2$ (coefficient 1) forms at 0.01 and NO$_2$ (coefficient 4) at 4 $	imes$ 0.01 = 0.04 mol dm$^{-3}$ s$^{-1}$.

**Step 1.** For $\mathrm{2\ N_2O_5(g) \longrightarrow 4\ NO_2(g) + O_2(g)}$:

$$	ext{rate} = -\dfrac{1}{2}\dfrac{\mathrm{d[N_2O_5]}}{\mathrm{d}t} = +\dfrac{1}{4}\dfrac{\mathrm{d[NO_2]}}{\mathrm{d}t} = +\dfrac{\mathrm{d[O_2]}}{\mathrm{d}t}$$

**Step 2 (rate of reaction).** Given $-\mathrm{d[N_2O_5]}/\mathrm{d}t$ = 0.02 mol dm$^{-3}$ s$^{-1}$: rate = $\frac{1}{2}$ $	imes$ 0.02 = 0.01 mol dm$^{-3}$ s$^{-1}$. …

Q.For the reaction $\mathrm{2\ N_2O_5(g) \longrightarrow 4\ NO_2(g) + O_2(g)}$ in liquid bromine, N $_2$ O $_5$ disappears at a rate of 0.02 moles dm $^{-3}$ sec $^{-1}$ . At what rate NO $_2$ and O $_2$ are formed? What would be the rate of reaction?

Concept understanding — Rate of Reaction

🎓Unlock everything free for 14 days

Unlock everything free for 14 days