Question

For the reaction: 4PH3(g) → P4(g) + 6H2(g) about 0.065 mol/s of PH3 is consumed in a 5.0 L flask. What are the rates of production of P4 and H2?

Answer 1

Answer: The rates of production of $P_4$ is $3.25\times 10^{-3}$ mol/Ls and $H_2$ is 0.0195 mol/Ls.

Explanation:

$4PH_3(g)\rightarrow P_4(g)+6H_2(g)$

Rate with respect to reactants is shown by negative sign as the reactants are decreasing with time and Rate with respect to products is shown by positive sign as the products are increasing with time.

Rate of the reaction=$-\frac{1}{4}\frac{[d[PH_3]}{dt}=\frac{[d[P_4]}{dt}=\frac{1}{6}\frac{[d[H_2]}{dt}$

Rate of decomposition of $PH_3=\frac{0.065 mol/s}{5.0L}=0.013mol/Ls$

Rate of production of $P_4=\frac{1}{4}\times {\text{rate of decomposition of}} PH_3=\frac{1}{4}\times 0.013=3.25\times 10^{-3}mol/Ls$

Rate of production of  $H_2=\frac{6}{4}\times {\text {rate of decomposition of}} PH_3=\frac{6}{4}\times 0.013=0.0195mol/Ls$

Answer 2

The rates of production of ${{\text{P}}_4}$ and ${{\text{H}}_{\text{2}}}$ are $0.00325{\text{ mol}} \cdot {{\text{L}}^{ - 1}} \cdot {{\text{s}}^{ - 1}}$ and $0.0195{\text{ mol}} \cdot {{\text{L}}^{ - 1}} \cdot {{\text{s}}^{ - 1}}$ respectively.

Further Explanation:

The rate of reaction is the pace at which reactants get converted into products. A general chemical reaction can be written as follows:

 ${\text{aA}} + {\text{bB}} \to {\text{cC}} + {\text{dD}}$

Here,

A and B are the reactants.

C and D are the products.

a and b are the stoichiometric coefficients of reactants.

c and d are the stoichiometric coefficients of products.

The expression for the rate of reaction for the above equation is as follows:

 ${\text{Rate of reaction}} = - \dfrac{1}{{\text{a}}}\dfrac{{d\left[ {\text{A}} \right]}}{{dt}} = - \dfrac{1}{{\text{b}}}\dfrac{{d\left[ {\text{B}} \right]}}{{dt}} = \dfrac{1}{{\text{c}}}\dfrac{{d\left[ {\text{C}} \right]}}{{dt}} = \dfrac{1}{{\text{d}}}\dfrac{{d\left[ {\text{D}} \right]}}{{dt}}$

Here, negative sign indicates the consumption of reactants whereas positive sign depicts the formation of products.

The given reaction is as follows:

 $4{\text{P}}{{\text{H}}_{\text{3}}}\left( g \right) \to {{\text{P}}_{\text{4}}}\left( g \right) + 6{{\text{H}}_{\text{2}}}\left( g \right)$

The expression for rate of reaction for the given reaction is as follows:

${\text{Rate of reaction}}= - \dfrac{1}{4}\dfrac{{d\left[ {{\text{P}}{{\text{H}}_{\text{3}}}} \right]}}{{dt}} = \dfrac{{d\left[ {{{\text{P}}_{\text{4}}}} \right]}}{{dt}} = \dfrac{1}{6}\dfrac{{d\left[ {{{\text{H}}_2}} \right]}}{{dt}}$                                  …… (1)

Here,

$\dfrac{{d\left[ {{\text{P}}{{\text{H}}_{\text{3}}}} \right]}}{{dt}}$ is the rate of consumption of ${\text{P}}{{\text{H}}_{\text{3}}}$.

$\dfrac{{d\left[ {{{\text{P}}_{\text{4}}}} \right]}}{{dt}}$ is the rate of formation of ${{\text{P}}_{\text{4}}}$.

$\dfrac{{d\left[ {{{\text{H}}_2}} \right]}}{{dt}}$ is the rate of formation of ${{\text{H}}_2}$.

The rate of decomposition of ${\text{P}}{{\text{H}}_{\text{3}}}$ can be determined as follows:

 $\begin{aligned}{\text{Rate of decomposition of P}}{{\text{H}}_{\text{3}}} &= \frac{{0.065{\text{ mol/s}}}}{{5{\text{ L}}}}\\&= 0.013{\text{ mol}} \cdot {{\text{L}}^{ - 1}} \cdot {{\text{s}}^{ - 1}}\\\end{aligned}$

The formula to calculate the rate of production of ${{\text{P}}_4}$ is as follows:

  ${\text{Rate of production of }}{{\text{P}}_4} = \frac{1}{4}\left( {{\text{Rate of decomposition of P}}{{\text{H}}_{\text{3}}}} \right)$                            …… (2)

The rate of decomposition of ${\text{P}}{{\text{H}}_{\text{3}}}$ is $0.013{\text{ mol}} \cdot {{\text{L}}^{ - 1}} \cdot {{\text{s}}^{ - 1}}$. Substitute this value in equation (2).

 $\begin{aligned}{\text{Rate of production of }}{{\text{P}}_4} &= \frac{1}{4}\left( {0.013{\text{ mol}} \cdot {{\text{L}}^{ - 1}} \cdot {{\text{s}}^{ - 1}}} \right)\\&= 0.00325{\text{ mol}} \cdot {{\text{L}}^{ - 1}} \cdot {{\text{s}}^{ - 1}}\\\end{aligned}$

Therefore the rate of production of ${{\text{P}}_{\text{4}}}$ is $0.00325{\text{ mol}} \cdot {{\text{L}}^{ - 1}} \cdot {{\text{s}}^{ - 1}}$.

The formula to calculate the rate of production of ${{\text{H}}_{\text{2}}}$ is as follows:

${\text{Rate of production of }}{{\text{H}}_{\text{2}}} = \dfrac{6}{4}\left( {{\text{Rate of decomposition of P}}{{\text{H}}_{\text{3}}}} \right)$             …… (3)

The rate of decomposition of ${\text{P}}{{\text{H}}_{\text{3}}}$ is $0.013{\text{ mol}} \cdot {{\text{L}}^{ - 1}} \cdot {{\text{s}}^{ - 1}}$. Substitute this value in equation (3).

$\begin{aligned}{\text{Rate of production of }}{{\text{H}}_{\text{2}}}&= \frac{6}{4}\left( {0.013{\text{ mol}} \cdot {{\text{L}}^{ - 1}} \cdot {{\text{s}}^{ - 1}}} \right)\\&= 0.0195{\text{ mol}} \cdot {{\text{L}}^{ - 1}} \cdot {{\text{s}}^{ - 1}} \\ \end{aligned}$

Therefore the rate of production of ${{\text{H}}_2}$ is $0.0195{\text{ mol}} \cdot {{\text{L}}^{ - 1}} \cdot {{\text{s}}^{ - 1}}$.

Learn more:

1. Rate of chemical reaction: brainly.com/question/1569924
2. The main purpose of conducting experiments: brainly.com/question/5096428

Answer Details:

Grade: Senior School

Chapter: Chemical Kinetics

Subject: Chemistry

Keywords: PH3, P4, H2, rate of formation, rate of decomposition, rate of reaction, 0.013 mol/L/s, 0.0195 mol/L/s, 0.00325 mol/L/s, rate of consumption, negative sign, positive sign.