Nitrogen gas N₂ and hydrogen gas H₂ combine to produce ammonia NH3 according to dy dX 1N₂ + 3H2 → 2NH3. Find r, and if ammonia is being produced at a rate of 5 moles dt dt per liter per hour. T = dY dt dX dt mol Lh mol Lh

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### Chemical Reactions and Rates: Reaction Rate Calculations

Chemical reactions produce one or more **product** substances from one or more **reactant** substances. If *x* molecules of reactant *X* combine with *y* molecules of reactant *Y* to produce *z* molecules of product *Z*, we write the reaction as:

\[ xX + yY \rightarrow zZ \]

The reaction rate \( r \) is the rate of change of concentration of one of the products over time. For this example, we can express the rate as:

\[ r = \frac{1}{z} \frac{dZ}{dt} = - \frac{1}{y} \frac{dY}{dt} = - \frac{1}{x} \frac{dX}{dt} \]

### Example Calculation

Consider the reaction where nitrogen gas \(\text{N}_2\) and hydrogen gas \(\text{H}_2\) combine to produce ammonia \(\text{NH}_3\) according to:

\[ \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \]

Find the rates \( r \), \( \frac{dY}{dt} \), and \( \frac{dX}{dt} \) if ammonia is being produced at a rate of 5 moles per liter per hour.

**Steps:**

1. Write the given rate of production of \( \text{NH}_3 \):
   \[ \frac{dZ}{dt} = 5 \text{ mol/Lh} \]

2. Calculate the reaction rate \( r \):
   \[ r = \frac{1}{2} \cdot 5 = 2.5 \text{ mol/Lh} \]

3. Using \( r \) to find \( \frac{dY}{dt} \):
   \[ r = \frac{1}{3} \left( -\frac{dY}{dt} \right) \]
   \[ -\frac{dY}{dt} = 3 \cdot 2.5 = 7.5 \text{ mol/Lh} \]

4. Using \( r \) to find \( \frac{dX}{dt} \):
   \[ -\frac{dX}{dt} = 2
Transcribed Image Text:### Chemical Reactions and Rates: Reaction Rate Calculations Chemical reactions produce one or more **product** substances from one or more **reactant** substances. If *x* molecules of reactant *X* combine with *y* molecules of reactant *Y* to produce *z* molecules of product *Z*, we write the reaction as: \[ xX + yY \rightarrow zZ \] The reaction rate \( r \) is the rate of change of concentration of one of the products over time. For this example, we can express the rate as: \[ r = \frac{1}{z} \frac{dZ}{dt} = - \frac{1}{y} \frac{dY}{dt} = - \frac{1}{x} \frac{dX}{dt} \] ### Example Calculation Consider the reaction where nitrogen gas \(\text{N}_2\) and hydrogen gas \(\text{H}_2\) combine to produce ammonia \(\text{NH}_3\) according to: \[ \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \] Find the rates \( r \), \( \frac{dY}{dt} \), and \( \frac{dX}{dt} \) if ammonia is being produced at a rate of 5 moles per liter per hour. **Steps:** 1. Write the given rate of production of \( \text{NH}_3 \): \[ \frac{dZ}{dt} = 5 \text{ mol/Lh} \] 2. Calculate the reaction rate \( r \): \[ r = \frac{1}{2} \cdot 5 = 2.5 \text{ mol/Lh} \] 3. Using \( r \) to find \( \frac{dY}{dt} \): \[ r = \frac{1}{3} \left( -\frac{dY}{dt} \right) \] \[ -\frac{dY}{dt} = 3 \cdot 2.5 = 7.5 \text{ mol/Lh} \] 4. Using \( r \) to find \( \frac{dX}{dt} \): \[ -\frac{dX}{dt} = 2
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