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3 years ago

Consider the reaction

Chemistry

2 answers:

spayn [35]3 years ago

7 0

Answer:

1- 0.04 M/s.

2- 0.16 M/s.

Explanation:

For the reaction: 4PH₃ → P₄ + 6H₂.

The rate of the reaction = - d[PH₃]/4dt = d[P₄]/dt = d[H₂]/6dt.

where, - d[PH₃]/dt is the rate of PH₃ changing "rate of disappearance of PH₃".

d[P₄]/dt is rate of P₄ changing "rate of appearance of P₄".

d[H₂]/dt is the rate of H₂ changing "rate of formation of H₂" (d[H₂]/dt = 0.24 M/s).

(a) At what rate is P₄ changing?

∵ The rate of the reaction = d[P₄]/dt = d[H₂]/6dt.

∴ rate of P₄ changing = d[P₄]/dt = d[H₂]/6dt = (0.240 M/s)/(6.0) = 0.04 M/s.

(b) At what rate is PH₃ changing?

∵ The rate of the reaction = - d[PH₃]/4dt = d[H₂]/6dt.

∴ rate of PH₃ changing = - d[PH₃]/dt = 4(d[H₂]/6dt) = (4)(0.240 M/s)/(6.0) = 0.16 M/s.

Ratling [72]3 years ago

7 0

a. P₄ is changing at rate : 0.04 M/s

b. PH₃ is changing at rate : -0.16 M/s

<h3>Further explanation </h3>

The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time.

Can be formulated:

Reaction: aA ---> bB

$\large{\boxed{\boxed{\bold{v~=~-\frac{\Delta A}{\Delta t}}}}$

$\large{\boxed{\boxed{\bold{v~=~+\frac{\Delta B}{\Delta t}}}}$

A = reagent

B = product

v = reaction rate

t = reaction time

For the reaction :

4PH₃(g) → P₄(g) + 6H₂(g)

Reaction rate is proportional with reaction coefficient :

$\displaystyle v=-\frac{1}{4}\frac{\Delta [PH_3]}{\Delta t}=+\frac{\Delta [P_4]}{\Delta t}=+\frac{1}{6}\frac{\Delta [H_2]}{\Delta t}$

Rate of H₂ = 0.24 M/s

then :

a. Rate of P₄ :

$\displaystyle v_{P_4}=\frac{1}{6}v_{H_2}\\\\v_{P_4}=\frac{1}{6}\times 0.24\\\\v_{P_4}=0.04~M/s$

b. Rate of PH₃ :

$\displaystyle v_{PH_3}=-\frac{4}{6}v_{H_2}\\\\v_{PH_3}=-\frac{4}{6}\times 0.24\\\\v_{P_4}=-0.16~M/s$

<h3>Learn more </h3>

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Keywords: reaction rate, molar concentration, products, reactants

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(1) The number of grams needed of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ are 19.23 g and 20.41 g respectively.

(2) The number of moles of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ are 0.641 moles and 0.352 moles respectively.

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$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

(4) The number of moles of $CO_2$ produced by burning each fuel is 1.28 mole and 1.41 mole respectively.

The fuel that emitting least amount of $CO_2$ is $C_2H_6$

Explanation :

Part 1 :

First we have to calculate the number of grams needed of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ .

As, 52 kJ energy required amount of $C_2H_6$ = 1 g

So, 1000 kJ energy required amount of $C_2H_6$ = $\frac{1000}{52}=19.23g$

and,

As, 49 kJ energy required amount of $C_4H_{10}$ = 1 g

So, 1000 kJ energy required amount of $C_4H_{10}$ = $\frac{1000}{49}=20.41g$

Part 2 :

Now we have to calculate the number of moles of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ .

Molar mass of $C_2H_6$ = 30 g/mole

Molar mass of $C_4H_{10}$ = 58 g/mole

$\text{ Moles of }C_2H_6=\frac{\text{ Mass of }C_2H_6}{\text{ Molar mass of }C_2H_6}=\frac{19.23g}{30g/mole}=0.641moles$

and,

$\text{ Moles of }C_4H_{10}=\frac{\text{ Mass of }C_4H_{10}}{\text{ Molar mass of }C_4H_{10}}=\frac{20.41g}{58g/mole}=0.352moles$

Part 3 :

Now we have to write down the balanced chemical equation for the combustion of the fuels.

The balanced chemical reaction for combustion of $C_2H_6$ is:

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

and,

The balanced chemical reaction for combustion of $C_4H_{10}$ is:

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

Part 4 :

Now we have to calculate the number of moles of $CO_2$ produced by burning each fuel to produce 1000 kJ.

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

From this we conclude that,

As, 1 mole of $C_2H_6$ react to produce 2 moles of $CO_2$

As, 0.641 mole of $C_2H_6$ react to produce $0.641\times 2=1.28$ moles of $CO_2$

and,

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

From this we conclude that,

As, 1 mole of $C_4H_{10}$ react to produce 4 moles of $CO_2$

As, 0.352 mole of $C_4H_{10}$ react to produce $0.352\times 4=1.41$ moles of $CO_2$

So, the fuel that emitting least amount of $CO_2$ is $C_2H_6$

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3 years ago