Chemistry, 19.03.2020 17:16 jadielmatmat

Consider this reaction:
2H3PO4(aq)→ P2O5(aq)+ 3H2O

At a certain temperature it obeys this rate law. rate
rate= (46.6M^-1. s^-1) [H3PO4]^2

Suppose a vessel contains H3PO4 at a concentration of 0.660M. Calculate how long it takes for the concentration of H3PO$ to decrease to 20% to its natural value. You may assume no other reaction is important. Round your answer to significant digits.

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Answer from: lidsrays

0.0345 sec

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

Rate=k[H_3PO_4]^2

k= rate constant = $46.6s^{-1}$

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process

for completion of 20 % of reaction

$t=\frac{2.303}{46.6}\log\frac{0.660}{\frac{20}{100}\times 0.660}$

$t=\frac{2.303}{46.6}\log\frac{0.660}{0.132}$

t=0.0345sec

The time taken for the concentration of H_3PO_4 to decrease to 20% to its natural value is 0.0345 sec

Answer from: Quest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.

Answer from: Quest

Aseasonal prevailing wind in the region of south and southeast asia, blowing from the southwest between may and september and bringing rain (the wet monsoon ), or from the northeast between october and april (the dry monsoon ).

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Consider this reaction:
2H3PO4(aq)→ P2O5(aq)+ 3H2O

At a certain temperature it ob...