Question

Assume that thermal decomposition of mercuric oxide, HgO, obeys first-order kinetics. It can be followed by the production of oxygen gas as a product: 2 HgO(s) --> 2 Hg(l) + O2 (g) At a particular temperature, k = 6.02 x 10-4 sec-1. If 1.00 gram of HgO were present initially, how long would it take to produce to produce 1 ml of O2 (g) at STP?

Answer 1

2 HgO(s) --> 2 Hg(l) + O₂ (g)

At STP, 1 mole of ideal gas occupy 22.4 L of volume.
Thus, number of moles in 1 mL of O₂(g) at STP = 1/(22.4*1000)
[O₂(g)] = 1/ 22400V

[HgO]₀ = n/V = 1 / 216.59V = 4.6170 * 10^-3 /V
From the reaction equation, number of moles left of HgO = (1 / 216.59) - 2*(1/ 22400) = 4.5277 * 10^-3
[HgO] = 4.5277 * 10^-3 /V

For 1st order reaction,
ln ([HgO]₀ / [HgO]) = kt
=> kt = 1.9531 * 10^-2
=> t = (1.9531 * 10^-2) / (6.02 * 10^-4) = 32.44 s