Question

1. the above is data for a reaction in which pb is a reactant.

a) calculate the overall rate of the reaction.
b) calculate the rate from 0 - 30s and from 30 - 60s.
c) explain why the 30 - 60s rate is lower than the 0 - 30s rate.

1. when caco3 reacts with hcl, co2(g) is produced. if 243ml of co2 is produced in 22s, what is the rate of the reaction in grams of caco3 per second?
2. a 5.0g sample of mg reacts completely with hcl in 120s. what is the average rate of this reaction in g mg per minute?

Explanation:

Step1: Calculate overall rate for Pb reaction

The overall rate of a reaction is given by $\text{Rate}=\frac{\Delta \text{mass}}{\Delta \text{time}}$. Initial mass of Pb is 65 g at $t = 0$ s and final mass is 25 g at $t=60$ s. So, $\text{Rate}=\frac{65 - 25}{60}=\frac{40}{60}\approx0.67$ g/s.

Step2: Calculate rate for 0 - 30s for Pb reaction

For the time - interval 0 - 30s, initial mass is 65 g and final mass is 41 g. $\text{Rate}=\frac{65 - 41}{30}=\frac{24}{30}=0.8$ g/s.

Step3: Calculate rate for 30 - 60s for Pb reaction

For the time - interval 30 - 60s, initial mass is 41 g and final mass is 25 g. $\text{Rate}=\frac{41 - 25}{30}=\frac{16}{30}\approx0.53$ g/s. The rate is lower in the 30 - 60s interval because as the reaction proceeds, the concentration of the reactant (Pb) decreases. According to the rate - law of chemical reactions, the rate of a reaction is often proportional to the concentration of reactants. As the amount of Pb decreases, the number of collisions between reactant particles decreases, leading to a lower reaction rate.

Step4: Calculate rate for $CaCO_3$ reaction

The balanced chemical equation is $CaCO_3 + 2HCl
ightarrow CO_2+H_2O + CaCl_2$. First, convert the volume of $CO_2$ to moles. At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 L. Given $V = 243$ mL $=0.243$ L. Moles of $CO_2,n_{CO_2}=\frac{0.243}{22.4}$ mol. From the stoichiometry of the reaction, 1 mole of $CaCO_3$ produces 1 mole of $CO_2$. Molar mass of $CaCO_3,M = 100.09$ g/mol. Mass of $CaCO_3$ reacted, $m_{CaCO_3}=n_{CO_2}\times100.09$. Rate of the reaction $r=\frac{m_{CaCO_3}}{22}$.
\[n_{CO_2}=\frac{0.243}{22.4}\text{mol}\]
\[m_{CaCO_3}=\frac{0.243}{22.4}\times100.09\text{g}\]
\[r=\frac{\frac{0.243}{22.4}\times100.09}{22}\approx0.049\text{g/s}\]

Step5: Calculate rate for Mg reaction

A 5.0 g sample of Mg reacts in 120 s. First, convert time to minutes. 120 s = 2 minutes. Rate of the reaction in g Mg per minute is $\frac{5.0}{2}=2.5$ g/min.

Answer:

1.
a. The overall rate of the Pb - reaction is approximately 0.67 g/s.
b. Rate from 0 - 30s is 0.8 g/s and from 30 - 60s is approximately 0.53 g/s.
c. The 30 - 60s rate is lower because the concentration of Pb (reactant) decreases as the reaction proceeds, reducing the number of collisions between reactant particles.

1. The rate of the $CaCO_3$ reaction is approximately 0.049 g/s.
2. The average rate of the Mg reaction is 2.5 g/min.