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inessss [21]
3 years ago
13

The chemical equation for the reaction of baking soda (sodium bicarbonate, nahco3) and vinegar (acetic acid, ch3cooh) may be wri

tten as two steps. fill in the missing information for the chemical equation. step 1: _____ + _____ → ch3coona(aq) + h2co3(l) step 2: h2co3(l) → h2o(l) + _____.
Chemistry
1 answer:
satela [25.4K]3 years ago
3 0
<h3><u>Answer;</u></h3>

Step 1;  NaHCO3(s) + CH3COOH(l)

Step 2 ; CO2(g)

<h3><u>Explanation;</u></h3>
  • The chemical equation for the reaction of baking soda (sodium bicarbonate, NaHCO3) and vinegar (acetic acid, CH3COOH) reaction occurs in two steps.

Step 1;  

  • A double displacement reaction in which acetic acid in the vinegar reacts with sodium bicarbonate to form sodium acetate and carbonic acid:
  • Equation;

NaHCO3(s)+ CH3COOH(l) → CH3COONa(aq) + H2CO3(l)

Step 2;

  • Carbonic acid is unstable and undergoes a decomposition reaction to produce the carbon dioxide gas:

H2CO3(l) → H2O(l) + CO2(g)

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3 years ago
The following mechanism has been suggested for the reaction between nitrogen monoxide and oxygen: NO(g) + NO(g) → N2O2(g) (fast)
Karo-lina-s [1.5K]

Answer:

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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3 years ago
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