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2 years ago

What is the correct equation for the reaction quotient of the following reaction? 2Fe3+(aq) + Zn(s) ⇌ 2Fe2+(aq) + Zn2+(aq)

Chemistry

1 answer:

Ira Lisetskai [31]2 years ago

3 0

Explanation:

A reaction quotient is defined as the ratio of concentration of products over reactants raised to the power of their stoichiometric coefficients.

A reaction quotient is denoted by the symbol Q.

For example, $2Fe^{3+}(aq) + Zn(s) \rightleftharpoons 2Fe^{2+}(aq) + Zn^{2+}(aq)$

The reaction quotient for this reaction is as follows.

Q = $\frac{[Fe^{2+}]^{2}[Zn^{2+}]}{[Fe^{3+}]^{2}}$

[Zn] will be equal to 1 as it is present in solid state. Therefore, we don't need to write it in the reaction quotient expression.

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2 years ago

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2 years ago

The gas-phase reaction follows an elementary rate law and is to be carried out first in a PFR and then in a separate experiment

astraxan [27]

Answer:

The activation energy is $=8.1\,kcal\,mol^{-1}$

Explanation:

The gas phase reaction is as follows.

$A \rightarrow B+C$

The rate law of the reaction is as follows.

$-r_{A}=kC_{A}$

The reaction is carried out first in the plug flow reactor with feed as pure reactant.

From the given,

Volume "V" = $10dm^{3}$

Temperature "T" = 300 K

Volumetric flow rate of the reaction $v_{o}=5dm^{3}s$

Conversion of the reaction "X" = 0.8

The rate constant of the reaction can be calculate by the following formua.

$V= \frac{v_{0}}{k}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]$

Rearrange the formula is as follows.

$k= \frac{v_{0}}{V}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]............(1)$

The feed has Pure A, mole fraction of A in feed $y_{A_{o}}$ is 1.

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacte.

$=1(2-1)=1$

Substitute the all given values in equation (1)

$k=\frac{5m^{3}/s}{10dm^{3}}[(1+1)ln \frac{1}{1-0.8}-1 \times 0.8] = 1.2s^{-1}$

Therefore, the rate constant in case of the plug flow reacor at 300K is $1.2s^{-1}$

The rate constant in case of the CSTR can be calculated by using the formula.

$\frac{V}{v_{0}}= \frac{X(1+\epsilon X)}{k(1-X)}.............(2)$

The feed has 50% A and 50% inerts.

Hence, the mole fraction of A in feed $y_{A_{o}}$ is 0.5

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacted.

$=0.5(2-1)=0.5$

Substitute the all values in formula (2)

$\frac{10dm^{3}}{5dm^{3}}=\frac{0.8(1+0.5(0.8))}{k(1-0.8)}=2.8s^{-1}$

Therefore, the rate constant in case of CSTR comes out to be $2.8s^{-1}$

The activation energy of the reaction can be calculated by using formula

$k(T_{2})=k(T_{1})exp[\frac{E}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})]$

In the above reaction rate constant at the two different temperatures.

Rearrange the above formula is as follows.

$E= R \times(\frac{T_{1}T_{2}}{T_{1}-T_{2}})ln\frac{k(T_{2})}{k(T_{1})}$

Substitute the all values.

$=1.987cal/molK(\frac{300K \times320K}{320K \times300K})ln \frac{2.8}{1.2}=8.081 \times10^{3}cal\,mol^{-1}$

$=8.1\,kcal\,mol^{-1}$

Therefore, the activation energy is $=8.1\,kcal\,mol^{-1}$

8 0

3 years ago

Consider the following equilibrium: CO2(g) + C(graphite) 2CO(g); ΔH = 172.5 kJ The equilibrium constant for this reaction will

bezimeni [28]

Answer: Option (B) is the correct answer.

Explanation:

As the given reaction is as follows.

$CO_{2}(g) + C(graphite) \rightarrow 2CO(g)$

Equilibrium constant for this reaction will be as follows.

$K_{c} = \frac{[CO_{2}]}{[CO]^{2}}$

According to Le Chatelier's principle, when we increase the temperature then the equilibrium will shift towards the right hand side.

As a result, concentration of carbon dioxide will decrease whereas concentration of carbon monoxide will increase.

Thus, we can conclude that in the given reaction equilibrium constant for this reaction will decrease with increasing temperature.

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3 years ago