4NH3(g) +50₂(g) = 4NO(g) + 6H₂O(g) AH° -1000. kJ K = 1.35 = The equilibrium concentrations are [NH3] = 0.20M, [02] = 0.90M, [H₂O] = 1.34 M, and [NO] = [?] What is the equilibrium concentration of NO at 25 °C? [NO] M Enter

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**Chemical Equilibrium Calculation**

Consider the following balanced chemical equation for the reaction:

\[ 4 \text{NH}_3(g) + 5 \text{O}_2(g) \rightleftharpoons 4 \text{NO(g)} + 6 \text{H}_2\text{O}(g) \]

The standard enthalpy change (\(\Delta H^\circ\)) for this reaction is \(-1000 \text{kJ}\), and the equilibrium constant (\(K\)) is given as \(1.35\).

**Given:**
- Equilibrium concentration of ammonia \([ \text{NH}_3 ] = 0.20 \text{M} \)
- Equilibrium concentration of oxygen \([ \text{O}_2 ] = 0.90 \text{M} \)
- Equilibrium concentration of water \([ \text{H}_2\text{O}] = 1.34 \text{M} \)

**Question:**
What is the equilibrium concentration of nitrogen monoxide \([ \text{NO}] \) at \(25^\circ \text{C}\)?

**Calculation:**

The equilibrium expression for the above reaction is:

\[ K = \frac{[\text{NO}]^4 [\text{H}_2\text{O}]^6}{[\text{NH}_3]^4 [\text{O}_2]^5} \]

Given \(K = 1.35\), we need to compute the concentration of \([ \text{NO} ]\):

\[ 1.35 = \frac{[\text{NO}]^4 (1.34)^6}{(0.20)^4 (0.90)^5} \]

Rearranging this equation to solve for \([ \text{NO} ]\):

\[[\text{NO}]^4 = 1.35 \cdot (0.20)^4 \cdot (0.90)^5 \div (1.34)^6\]

For simplicity, let us denote the reaction quotient without powers as follows:
Let \( a = (0.20)^4 \)
Let \( b = (0.90)^5 \)
Let \( c = (1.34)^6 \)

Plug in values:
\[ a = (0.20)^4 = 0.0016 \]
\[ b = (0.90
Transcribed Image Text:**Chemical Equilibrium Calculation** Consider the following balanced chemical equation for the reaction: \[ 4 \text{NH}_3(g) + 5 \text{O}_2(g) \rightleftharpoons 4 \text{NO(g)} + 6 \text{H}_2\text{O}(g) \] The standard enthalpy change (\(\Delta H^\circ\)) for this reaction is \(-1000 \text{kJ}\), and the equilibrium constant (\(K\)) is given as \(1.35\). **Given:** - Equilibrium concentration of ammonia \([ \text{NH}_3 ] = 0.20 \text{M} \) - Equilibrium concentration of oxygen \([ \text{O}_2 ] = 0.90 \text{M} \) - Equilibrium concentration of water \([ \text{H}_2\text{O}] = 1.34 \text{M} \) **Question:** What is the equilibrium concentration of nitrogen monoxide \([ \text{NO}] \) at \(25^\circ \text{C}\)? **Calculation:** The equilibrium expression for the above reaction is: \[ K = \frac{[\text{NO}]^4 [\text{H}_2\text{O}]^6}{[\text{NH}_3]^4 [\text{O}_2]^5} \] Given \(K = 1.35\), we need to compute the concentration of \([ \text{NO} ]\): \[ 1.35 = \frac{[\text{NO}]^4 (1.34)^6}{(0.20)^4 (0.90)^5} \] Rearranging this equation to solve for \([ \text{NO} ]\): \[[\text{NO}]^4 = 1.35 \cdot (0.20)^4 \cdot (0.90)^5 \div (1.34)^6\] For simplicity, let us denote the reaction quotient without powers as follows: Let \( a = (0.20)^4 \) Let \( b = (0.90)^5 \) Let \( c = (1.34)^6 \) Plug in values: \[ a = (0.20)^4 = 0.0016 \] \[ b = (0.90
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