of students made 5 standard solutions and measured their corresponding absorbance values at group a wavelength of 525.0 nm to generate the graph below. Absorbance of Standard Solutions to Find Concentration of Unknown X Solution 0.50 y = 75.9x + 0.0045 R? = 0.9946 0.40 0.30 0.20 0.10 0.00 0.001 0.002 0.003 0.004 0.005 0.006 Concentration of X (mole/L) The colorless reactants V and W form unknown X from the reaction: 2V + W S X A solution is prepared by mixing together the following: Volume 0.0150 M V (mL) Volume 0.0150 M W (mL) 6.00 4.00 This solution reaches equilibrium, and then is placed in a 1.00 cm wide cuvet and inserted into the spectrometer, producing an absorbance reading of 0.275 at a wavelength of 525.0 nm. Calculate Keg for the above reaction, including units. Absorbance

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### Determination of Equilibrium Constant \( K_{eq} \)

#### Experimental Background
A group of students prepared 5 standard solutions and measured their corresponding absorbance values at a wavelength of 525.0 nm. These values were then used to generate the calibration graph illustrated below.

#### Calibration Graph
**Title:** Absorbance of Standard Solutions to Find Concentration of Unknown X Solution

![Graph](https://example.com/image.png)  <!-- Note: Replace with real image link -->

**Description:**
- **X-axis:** Concentration of X (mole/L)
- **Y-axis:** Absorbance
- **Best Fit Line Equation:** \( y = 75.9x + 0.0045 \)
- **R-squared Value:** \( R^2 = 0.9946 \)

The graph shows a linear relationship between the concentration of X (mole/L) and the absorbance, with the line equation given above.

#### Reaction
The colorless reactants V and W form unknown X from the reaction:

\[ 2V + W \rightleftharpoons X \]

#### Preparation of Solution
A solution was prepared by mixing the following volumes:

| Volume 0.0150 M V (mL) | Volume 0.0150 M W (mL) |
|------------------------|------------------------|
| 6.00                   | 4.00                   |

#### Experimental Data
This solution reached equilibrium and was then placed in a 1.00 cm wide cuvet for measurement. The spectrometer provided an absorbance reading of 0.275 at a wavelength of 525.0 nm.

#### Calculation of \( K_{eq} \)

1. **Calculate the concentration of X using the calibration curve:**

   Given:
   \[ \text{Absorbance (A)} = 0.275 \]
   Using the best fit line equation:
   \[ A = 75.9 \cdot [X] + 0.0045 \]

   Solving for \([X]\):
   \[ 0.275 = 75.9 \cdot [X] + 0.0045 \]
   \[ 75.9 \cdot [X] = 0.275 - 0.0045 \]
   \[ 75.9 \cdot [X] = 0.2705 \]
   \
Transcribed Image Text:### Determination of Equilibrium Constant \( K_{eq} \) #### Experimental Background A group of students prepared 5 standard solutions and measured their corresponding absorbance values at a wavelength of 525.0 nm. These values were then used to generate the calibration graph illustrated below. #### Calibration Graph **Title:** Absorbance of Standard Solutions to Find Concentration of Unknown X Solution ![Graph](https://example.com/image.png) <!-- Note: Replace with real image link --> **Description:** - **X-axis:** Concentration of X (mole/L) - **Y-axis:** Absorbance - **Best Fit Line Equation:** \( y = 75.9x + 0.0045 \) - **R-squared Value:** \( R^2 = 0.9946 \) The graph shows a linear relationship between the concentration of X (mole/L) and the absorbance, with the line equation given above. #### Reaction The colorless reactants V and W form unknown X from the reaction: \[ 2V + W \rightleftharpoons X \] #### Preparation of Solution A solution was prepared by mixing the following volumes: | Volume 0.0150 M V (mL) | Volume 0.0150 M W (mL) | |------------------------|------------------------| | 6.00 | 4.00 | #### Experimental Data This solution reached equilibrium and was then placed in a 1.00 cm wide cuvet for measurement. The spectrometer provided an absorbance reading of 0.275 at a wavelength of 525.0 nm. #### Calculation of \( K_{eq} \) 1. **Calculate the concentration of X using the calibration curve:** Given: \[ \text{Absorbance (A)} = 0.275 \] Using the best fit line equation: \[ A = 75.9 \cdot [X] + 0.0045 \] Solving for \([X]\): \[ 0.275 = 75.9 \cdot [X] + 0.0045 \] \[ 75.9 \cdot [X] = 0.275 - 0.0045 \] \[ 75.9 \cdot [X] = 0.2705 \] \
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