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loris [4]
3 years ago
11

Illustrate the x- and y-axes to show the reaction pathway and potential energy, in kilojoules. Ensure your energy intervals are

appropriate for the data.
Plot the enthalpy values of the reactants, products, and transition state using three horizontal dotted lines across the graph for each.

Draw the energy curve from the reactants line to the transition state and curve the line back down to the energy of the products. Label the reactants, products, and transition state.

Illustrate double-headed arrows to represent both the total change in enthalpy (ΔH) and the activation energy (Ea).

Calculate the total change in enthalpy and the activation energy using the energy values provided for each reaction. Record those values below the graph.

Make sure correct units are included.

I know this is a lot, so I'm giving more points than I ever have. Thank you!!!

Chemistry
1 answer:
Rufina [12.5K]3 years ago
6 0

Answer:

Enthalpy of a reaction = P - R  

Synthesis: ΔH = 25 kJ - (-30 kJ) = 55 kJ  

Single Replacement: ΔH = 30kJ - 65 kJ = -35 kJ  

Double Replacement: ΔH = 60 kJ - 10 kJ = 50 kJ

Explanation:

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9. Given the following reaction:CO (g) + 2 H2(g) CH3OH (g)In an experiment, 0.45 mol of CO and 0.57 mol of H2 were placed in a 1
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Answer:

Keq=11.5

Explanation:

Hello,

In this case, for the given reaction at equilibrium:

CO (g) + 2 H_2(g) \rightleftharpoons CH_3OH (g)

We can write the law of mass action as:

Keq=\frac{[CH_3OH]}{[CO][H_2]^2}

That in terms of the change x due to the reaction extent we can write:

Keq=\frac{x}{([CO]_0-x)([H_2]_0-2x)^2}

Nevertheless, for the carbon monoxide, we can directly compute x as shown below:

[CO]_0=\frac{0.45mol}{1.00L}=0.45M\\

[H_2]_0=\frac{0.57mol}{1.00L}=0.57M\\

[CO]_{eq}=\frac{0.28mol}{1.00L}=0.28M\\

x=[CO]_0-[CO]_{eq}=0.45M-0.28M=0.17M

Finally, we can compute the equilibrium constant:

Keq=\frac{0.17M}{(0.45M-0.17M)(0.57M-2*0.17M)^2}\\\\Keq=11.5

Best regards.

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