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Katarina [22]
3 years ago
6

Be sure to answer all parts. identify and label the species in each reaction. (a) nh4+(aq) + h2o(l) ⇌ nh3(aq) + h3o+(aq) acid ba

se acid base conjugate acid conjugate base conjugate acid conjugate base (b) cn−(aq) + h2o(l) ⇌ hcn(aq) + oh−(aq) acid base acid base conjugate acid conjugate base conjugate acid conjugate base
Chemistry
1 answer:
quester [9]3 years ago
5 0
A)

NH⁴⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃0⁺<span>(aq)

- acid </span>a species that able to donate (H+): NH⁴⁺
- base a species that is able to accept a proton (H+): H₂O
- conjugate base a species formed when acid donates a proton (H+): NH₃
- conjugate acid a species formed by a base accepts a proton (H+): H₃0⁺

b)

CN⁻(aq) + H₂O(l) ⇌ HCN(aq) + OH⁻(aq)

- base a species that is able to accept a proton (H+): CN⁻
- acid a species that able to donate (H+): H₂O
- conjugate acid a species formed by a base accepts a proton (H+): HCN
- conjugate base a species formed when acid donates a proton (H+): OH⁻

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The coefficients are 2 for H₂O and 1 for Ca(OH)₂.

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Ca(OH)₂(aq) + 2 HCl(aq) → CaCl₂(aq) + 2 H₂O(l)

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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 is1.2s^{-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
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