Chemistry, 25.06.2019 14:30 gissell30

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?

Answers: 1

Show answers

Answers

Answer from: seymani2

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:

Rate of chemical reaction: The main purpose of conducting experiments:

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.

Answer from: Quest

Iwould you figure it out but you don't have a picture or in your question as to what the physical property you're looking for is

Answer from: Quest

answerlinnaeus;