Question

A chemistry graduate student is studying the rate of this reaction: 2H3PO4 (aq) ---> P2O5 (aq) + 3H2O (aq)He fills a reaction vessel with and measures its concentration as the reaction proceeds: time (seconds) [H3PO4]0 0.03M1 0.018M2 0.011M3 0.0067M4 0.0041MUse this data to answer the following questions.Write the rate law of the reaction.

Answer 1

Answer:

The rate law of the reaction will be;

$R=k[H_3PO_4]^1$

Explanation:

$2H_3PO_4 (aq)\rightarrow P_2O_5 (aq) + 3H_2O (aq)$

Let the rate law of the reaction be :$R=k[H_3PO_4]^a$

Where : R is rate of the reaction at at given concentration of $H_3PO_4$ and k is rate constant.

Rate of reaction from zero second to 1 second:

$R=-\frac{0.018M-0.03M}{1-0}=0.012 M/s$

$0.012 M/s=k[0.018 M]^a$..[1]

Rate of reaction from 1 second to 2 second:

$R=-\frac{0.011 M-0.018M}{2-1}=0.007 M/s$

$0.007 M/s=k[0.011 M]^a$..[2]

[1] ÷ [2]

$\frac{0.012 M/s}{0.007M/s}=\frac{k[0.018 M]^a}{k[0.011 M]^a}$

a = 1.09 ≈ 1

The rate law of the reaction will be;

$R=k[H_3PO_4]^1$