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Brrunno [24]
3 years ago
14

Given only the following data, what can be said about the following reaction?3H2(g) + N2(g)---> 2NH3(g) ΔH=-92kJA.) The entha

lpy of products is greater than the enthalpy of reactantsB.) The total bond energies of products are greater than the total bond energies of reactantsC.) The reaction is very fastD.) Nitrogen and Hydrogen are very stable bonds compared to the bonds of ammonia
Chemistry
1 answer:
emmainna [20.7K]3 years ago
3 0

Answer:

D.) Nitrogen and Hydrogen are very stable bonds compared to the bonds of ammonia.

Explanation:

For the reaction:

3H₂(g) + N₂(g) → 2NH₃(g)

The enthalpy change is ΔH = -92kJ

This enthalpy change is defined as the enthalpy of products - the enthalpy of reactants. As the enthalpy is <0, The enthalpy of products is <em>lower </em>than the enthalpy of reactants.

Also, it is possible to obtain the enthalpy change from the bond energies of products - bond energies of reactants, thus, The total bond energies of products are <em>lower</em> than the total bond energies of reactants.

The rate of the reaction couldn't be determined using ΔH.

As the bond energy of ammonia is lower than bonds of nitrogen and hydrogen, <em>D. Nitrogen and Hydrogen are very stable bonds compared to the bonds of ammonia.</em>

I hope it helps!

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Which statement is true with respect to standard reduction potentials?
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C). Half-reactions with SRP values greater than zero are spontaneous.

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How many kilocalories are needed to vaporize 5.8 mol of Br2
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For the reaction: 2 A+B 3 C the rate of disappearance of B was 0.30 M/s. What is the rate of disappearance of A and the rate of
GenaCL600 [577]

Answer:

rA = 0.60 M/s

rC = 0.90 M/s

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2 A+B ⇒ 3 C

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rA=-\frac{\Delta[A] }{\Delta t}

rB=-\frac{\Delta[B] }{\Delta t}

rC=\frac{\Delta[C] }{\Delta t}

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r= \frac{rA}{2} =\frac{rB}{1} =\frac{rC}{3}

The rate of disappearance of B is 0.30 M/s.

The rate of disappearance of A is:

\frac{rA}{2} =\frac{rB}{1}\\rA = 2 \times rB = 2 \times 0.30 M/s = 0.60 M/s

The rate of appearance of C is:

\frac{rB}{1} =\frac{rC}{3}\\rC = 3 \times rB = 3 \times 0.30 M/s = 0.90 M/s

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