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Consider the reaction: 2NO(g)+O(2)(g) ...

Consider the reaction:
`2NO(g)+O_(2)(g) rarr 2NO_(2)(g)`
Calculated the standard Gibbs energy change at `298K` and predict whether the reaction is spontaneous or not. `Delta_(f)G^(Theta) (NO) = 86.69 kJ mol^(-1), Delta_(f)G^(Theta) (NO_(2)) = 51.84 kJ mol^(-1)`.

A

Yes, spontaneous

B

No, the reaction is Non-spontaneous

C

Equilibrium

D

cannot predict

Text Solution

AI Generated Solution

The correct Answer is:
To calculate the standard Gibbs energy change (ΔG) for the reaction: \[ 2NO(g) + O_2(g) \rightarrow 2NO_2(g) \] we will use the formula: \[ \Delta G = \Delta G_{f}^{\Theta} \text{(products)} - \Delta G_{f}^{\Theta} \text{(reactants)} \] ### Step 1: Identify the Gibbs free energy of formation values From the question, we have the following values: - \( \Delta_{f}G^{\Theta} (NO) = 86.69 \, \text{kJ mol}^{-1} \) - \( \Delta_{f}G^{\Theta} (NO_2) = 51.84 \, \text{kJ mol}^{-1} \) ### Step 2: Calculate the Gibbs free energy for the products Since there are 2 moles of \( NO_2 \) produced, we calculate: \[ \Delta G_{f}^{\Theta} \text{(products)} = 2 \times \Delta_{f}G^{\Theta} (NO_2) \] \[ = 2 \times 51.84 \, \text{kJ mol}^{-1} \] \[ = 103.68 \, \text{kJ} \] ### Step 3: Calculate the Gibbs free energy for the reactants Since there are 2 moles of \( NO \) consumed, we calculate: \[ \Delta G_{f}^{\Theta} \text{(reactants)} = 2 \times \Delta_{f}G^{\Theta} (NO) \] \[ = 2 \times 86.69 \, \text{kJ mol}^{-1} \] \[ = 173.38 \, \text{kJ} \] ### Step 4: Calculate the standard Gibbs energy change (ΔG) Now we can substitute the values into the Gibbs energy change formula: \[ \Delta G = \Delta G_{f}^{\Theta} \text{(products)} - \Delta G_{f}^{\Theta} \text{(reactants)} \] \[ = 103.68 \, \text{kJ} - 173.38 \, \text{kJ} \] \[ = -69.70 \, \text{kJ} \] ### Step 5: Determine spontaneity A reaction is spontaneous if \( \Delta G < 0 \). Since we calculated \( \Delta G = -69.70 \, \text{kJ} \), which is negative, we conclude that the reaction is spontaneous. ### Final Answer The standard Gibbs energy change at 298 K is \( -69.70 \, \text{kJ} \), and the reaction is spontaneous. ---
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