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kykrilka [37]

What? I don’t think anyone knows what that is

7 0

3 years ago

Sulfonation of benzene has the following mechanism:(b) Write the overall rate law in terms of the initial rate of the reaction.

erica [24]

The overall rate law in terms of the initial rate of the reaction is:

= $k_{overall} [H_{2} SO_{4} ]^{2} [C_{6}H_{6}]$

The Sulfonation of benzene has the following mechanism:

(1) $2H_{2} SO_{4}$ ⇌ $H_{3}O^{+} + HSO_{4} + SO_{3}$

[fast]

(2) $SO_{3} + C_{6} H_{6}$ → $H(C_{6} H_{5}^{+} )SO_{3}^{-}$

[slow]

(3) $H(C_{6} H_{5}^{+} )SO_{3}^{-}$ $+ HSO_{4}^{-}$ → $C_{6} H_{5} SO_{3}^{-} + H_{2} SO_{4}$

[fast]

(4) $C_{6} H_{5} SO_{3}^{-} + H_{3}O^{+}$ → $C_{6} H_{5} SO_{3} H + H_{2} O$

[fast]

<h3>Calculation of rate law:</h3>

Given that the reaction's slowest step is the rate-dependent step, the following is the definition of rate law:

Rate = $k_{2} [SO_{3} ][C_{6} H_{6} ]$

However, the issue is that because $SO_{3}$ is an intermediate, it cannot be accounted for in the general rate legislation.

The following rate rule governs the synthesis of $SO_{3}$ :

Rate = $k_{1}[H_{2} SO_{4} ]^{2}$

Consequently, if we replace equation 1 with equation 2,

Rate is determined to be = $k_{overall} [H_{2} SO_{4} ]^{2} [C_{6}H_{6}]$

Hence, the overall rate law in terms of the initial rate of the reaction is:

= $k_{overall} [H_{2} SO_{4} ]^{2} [C_{6}H_{6}]$

Learn more about rate law here:

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6 0

1 year ago

Is krypton ionic or covalence

allochka39001 [22]

Krypton is covalence

5 0

3 years ago

Read 2 more answers

Do Alkali Metal Compounds form precipitates? And Why?

Alexeev081 [22]

Answer:

Alkali metal hydroxides can be used to test the identity of metals in certain salts. The colour of the precipitate will help identify the metal : Calcium hydroxide is soluble; no precipitate is formed.

6 0

3 years ago

One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

<u />

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

3 years ago