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Naya [18.7K]
4 years ago
12

Urea is an organic compound widely used as a fertilizer. Its solubility in water allows it to be made into aqueous fertilizer so

lutions and applied to crops in a spray.
H2N-CO-NH2

What is the maximum theoretical number of water molecules with which one urea molecule can hydrogen bond? (Exact whole number required.)

Chemistry
1 answer:
stiks02 [169]4 years ago
4 0

Answer:

One urea molecule can make hydrogen bonding with 8 water molecules

Explanation:

  • Hydrogen bonding takes place between lone pair of electronegative atoms (O,N and F) and H atom attached with those electronegative atoms.
  • Draw the lewis structure of urea and determine how many lone pairs and H atoms are available for hydrogen bonding formation.
  • Theoretically one urea can make hydrogen bonding with 8 water molecules.
  • Representation of hydrogen bonding has been given below.

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When a substance such as ice melts, its temperature increases. Describe what happens to the arrangement of the water molecules a
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The following mechanism has been suggested for the reaction between nitrogen monoxide and oxygen: NO(g) + NO(g) → N2O2(g) (fast)
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b. Second order in NO and first order in O₂.

Explanation:

A. The mechanism

\rm 2NO\xrightarrow[k_{-1}]{k_{1}}N_{2}O_{2} \, (fast)\\\rm N_{2}O_{2} + O_{2}\xrightarrow{k_{2}} 2NO_{2} \, (slow)

B. The rate expressions

-\dfrac{\text{d[NO]} }{\text{d}t} = k_{1}[\text{NO]}^{2} - k_{-1} [\text{N}_{2}\text{O}_{2}]^{2}\\\\\rm -\dfrac{\text{d[N$_{2}$O$_{2}$]}}{\text{d}t} = -\dfrac{\text{d[O$_{2}$]}}{\text{d}t} = k_{2}[ N_{2}O_{2}][O_{2}] - k_{1} [NO]^{2}\\\\\dfrac{\text{d[NO$_{2}$]}}{\text{d}t}= k_{2}[ N_{2}O_{2}][O_{2}]

The last expression is the rate law for the slow step. However, it contains the intermediate N₂O₂, so it can't be the final answer.

C. Assume the first step is an equilibrium

If the first step is an equilibrium, the rates of the forward and reverse reactions are equal. The equilibrium is only slightly perturbed by the slow leaking away of N₂O₂ to form product.

\rm k_{1}[NO]^{2} = k_{-1} [N_{2}O_{2}]\\\\\rm [N_{2}O_{2}] = \dfrac{k_{1}}{k_{-1}}[NO]^{2}

D. Substitute this concentration into the rate law

\rm \dfrac{\text{d[NO$_{2}$]}}{\text{d}t}= \dfrac{k_{2}k_{1}}{k_{-1}}[NO]^{2} [O_{2}] = k[NO]^{2} [O_{2}]

The reaction is second order in NO and first order in O₂.

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