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bagirrra123
2 months ago
6

(a) The original value of the reaction quotient, Qc, for the reaction of H2(g) and I2(g) to form HI(g) (before any reactions tak

e place and before equilibrium is established), was 5.56. On the following graph, plot the points representing the initial concentrations of all three gases. Label each point with the formula of the gas.

Chemistry
1 answer:
KiRa [2.9K]2 months ago
7 0

Response:

Here's my calculation

Clarification:

Assume the starting concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.

We need to determine the initial concentration of HI.

1. We will need a chemical equation with concentrations, so let's compile all the information in one location.

H₂ + I₂ ⇌ 2HI

I/mol·L⁻¹: 0.30 0.15 x

2. Calculate the concentration of HI

Q_{\text{c}} = \dfrac{\text{[HI]}^{2}} {\text{[H$_{2}$][I$_{2}$]}} =\dfrac{x^{2}}{0.30 \times 0.15} =  5.56\\\\x^{2} = 0.30 \times 0.15 \times 5.56 = 0.250\\x = \sqrt{0.250} = \textbf{0.50 mol/L}\\\text{The initial concentration of HI is $\large \boxed{\textbf{0.50 mol/L}}$}

3. Plot the initial values

The graph below visualizes the initial concentrations as plotted on the vertical axis.

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Answer:

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For B: Free energy change is expressed in kJ/mol

For C: The forward reaction favors progression, while the reverse reaction does not.

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Regarding the specified chemical reaction:

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The relationship between standard Gibbs free energy and equilibrium constant is as follows:

\Delta G^o=-RT\ln K_{eq}

The free energy change can be calculated using the following equation:

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