6. The water gas shift reaction shown below can be used to produce hydrogen from water and coal. Given the data in the following table, which experiment will proceed in the forward direction to produce more hydrogen? Under the experimental conditions, Ke = 25 for this reaction. H2O(g) + CO(g) 5 H:(g) + CO2(g) Initial Concentrations (mol/L) Experiment [H;O] [CO] 0.030 [CO] [H2] I 0.030 II 0.030 0.030 III 0.020 0.020 0.050 0.040 IV 0.030 0.020 0.030 0.020 I only b. II only I and III d. I, III, and IV e. I and IV а. с.

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**Water Gas Shift Reaction: Producing Hydrogen from Water and Coal**

The water gas shift reaction, represented by the equation below, can be used to generate hydrogen:

\[ \text{H}_2\text{O(g)} + \text{CO(g)} \rightleftharpoons \text{H}_2\text{(g)} + \text{CO}_2\text{(g)} \]

Given the experimental conditions and the equilibrium constant (\( K_c = 25 \)), we can determine which experiment will proceed in the forward direction to produce more hydrogen. The initial concentrations (in mol/L) of the reactants and products for each experiment are provided in the table below:

**Initial Concentrations (mol/L)**

| Experiment | [H₂O] | [CO]  | [H₂]  | [CO₂] |
|------------|-------|-------|-------|-------|
| I          | 0.030 | 0.030 | 0     | 0     |
| II         | 0     | 0     | 0.030 | 0.030 |
| III        | 0.020 | 0.020 | 0.050 | 0.040 |
| IV         | 0.030 | 0.020 | 0.030 | 0.020 |

**Question:** Which experiment(s) will proceed in the forward direction to produce more hydrogen?

**Options:**
a. I only  
b. II only  
c. I and III  
d. I, III, and IV  
e. I and IV  

To answer this question, we compare the initial reaction quotient, \( Q_c \), to the equilibrium constant, \( K_c \). The reaction will proceed in the forward direction if \( Q_c < K_c \).

For each experiment, \( Q_c \) is calculated using:
\[ Q_c = \frac{[\text{H}_2][\text{CO}_2]}{[\text{H}_2\text{O}][\text{CO}]} \]

By evaluating each option and considering the given \( K_c \), we can determine that:

- In Experiment I:
  \[ Q_c = \frac{(0)(0)}{(0.030)(0.030)} = 0 \]
  Since \( 0 < 25 \), Experiment I will proceed in the
Transcribed Image Text:**Water Gas Shift Reaction: Producing Hydrogen from Water and Coal** The water gas shift reaction, represented by the equation below, can be used to generate hydrogen: \[ \text{H}_2\text{O(g)} + \text{CO(g)} \rightleftharpoons \text{H}_2\text{(g)} + \text{CO}_2\text{(g)} \] Given the experimental conditions and the equilibrium constant (\( K_c = 25 \)), we can determine which experiment will proceed in the forward direction to produce more hydrogen. The initial concentrations (in mol/L) of the reactants and products for each experiment are provided in the table below: **Initial Concentrations (mol/L)** | Experiment | [H₂O] | [CO] | [H₂] | [CO₂] | |------------|-------|-------|-------|-------| | I | 0.030 | 0.030 | 0 | 0 | | II | 0 | 0 | 0.030 | 0.030 | | III | 0.020 | 0.020 | 0.050 | 0.040 | | IV | 0.030 | 0.020 | 0.030 | 0.020 | **Question:** Which experiment(s) will proceed in the forward direction to produce more hydrogen? **Options:** a. I only b. II only c. I and III d. I, III, and IV e. I and IV To answer this question, we compare the initial reaction quotient, \( Q_c \), to the equilibrium constant, \( K_c \). The reaction will proceed in the forward direction if \( Q_c < K_c \). For each experiment, \( Q_c \) is calculated using: \[ Q_c = \frac{[\text{H}_2][\text{CO}_2]}{[\text{H}_2\text{O}][\text{CO}]} \] By evaluating each option and considering the given \( K_c \), we can determine that: - In Experiment I: \[ Q_c = \frac{(0)(0)}{(0.030)(0.030)} = 0 \] Since \( 0 < 25 \), Experiment I will proceed in the
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