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lina2011 [118]
2 years ago
9

If sulfur dioxide and oxygen can be made into sulfur trioxide, what is the reverse reaction? A. SO3 +O2à SO3 B. 2SO2 + O2à 2SO3

C. 2SO3à2SO2+O2. D. SO2+2SO3à 3S +4O2
Chemistry
2 answers:
ycow [4]2 years ago
7 0
REVERSE REACTION WILL BE SO2+O2
solong [7]2 years ago
3 0

<u>Answer: </u>The correct answer is Option C.

<u>Explanation:</u>

Reverse reaction is defined as the chemical reaction in which the reactants forming products at first, react together to give back the reactants.

Forward reaction is given by the equation:

A+B\rightarrow C+D

Now, the reverse reaction for this will be:

C+D\rightarrow A+B

  • According to the question:

Sulfur dioxide and oxygen are reacting together to form sulfur trioxide. The forward reaction for this will be:

2SO_2+O_2\rightarrow 2SO_3

The reverse reaction for this will be:

2SO_2\rightarrow 2SO_2+O_2

Hence, the correct answer is Option C.

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What is the product of reduction of ethyl 4-oxobutanoate with sodium borohydride in ethanol at room temperature for 30 minutes?
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The product of reduction of ethyl 4-oxobutanoate with sodium borohydride in ethanol at room temperature for 30 minutes is ethyl 4- hydroxybutanoate .

Sodium borohydride is a relatively selective reducing agent  Ethanolic solutions of Sodium borohydride reduces aldehyde , and ketone , in the presence of acid chloride , ester , epoxide , lactones , acids , nitriles , nitro groups.

The sodium borohydride does not reduce ester group because sodium borohydride is not strong enough and the electrophilicity at carbony carbon of ester is not more as compare toaldehyde , and ketone

The product of reduction of ethyl 4-oxobutanoate with sodium borohydride in ethanol at room temperature for 30 minutes is ethyl 4- hydroxybutanoate .

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Use bond energies from Table 10.3 in the textbook to estimate the enthalpy change (ΔH) for the following reaction. C2H2(g)+H2(g)
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Answer: =176.6kJmol^{-1}

Explanation:Bond energy of H-H is 436.4 kJ/mole

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Bond energy of C=C is 620 kJ/mol

Bond energy of C≡C is 835 kJ/mol

\Delta H= {\text {sum of bond energies of reactants}}-  {\text {sum of bond energies of products}}

\Delta H= {1B.E(C≡C)+2B.E(C-H) +1B.E(H-H)} - {1B.E(C=C)+4B.E(C-H)}

\Delta H= {1B.E(835kJmole^{-1})+2B.E(414kJmole^{-1}) +1B.E(436.4kJmole^{-1})} -  {1B.E(620kJmole^{-1})+4B.E(414kjmole^{-1})}

=176.6kJmol^{-1}







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