For the reaction \( \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \) at 298 K, the enthalpy change \( \Delta H = -92.4 \, \text{kJ/mol} \). What happens to the equilibrium when temperature is increased?
For exothermic reactions (\( \Delta H < 0 \)), increasing temperature shifts the equilibrium towards the reactants (left), as per Le Chatelier’s principle.
- The reaction is: \( \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \), with \( \Delta H = -92.4 \, \text{kJ/mol} \).
- Since \( \Delta H < 0 \), the reaction is exothermic (releases heat).
- According to Le Chatelier’s principle, for an exothermic reaction, increasing the temperature favors the endothermic direction (reverse reaction) to absorb the added heat.
- Here, the reverse reaction is: \( 2\text{NH}_3(g) \rightarrow \text{N}_2(g) + 3\text{H}_2(g) \), which is endothermic.
- Thus, increasing the temperature shifts the equilibrium to the left (towards reactants).
- This matches option (B).
At 700 K, the equilibrium constant $K_e$ for the reaction $ \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) $ is 0.2 mol L$^{-2}$. What is the value of $K$ for the reverse reaction?
For the reaction A(g) $\rightleftharpoons$ 2B(g), the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500, at 1000 K.
[Given: R = 0.0831 atm $mol^{–1} K^{–1}$]
$K_p$ for the reaction at 1000 K is:
The equilibrium constant for decomposition of $ H_2O $ (g) $ H_2O(g) \rightleftharpoons H_2(g) + \frac{1}{2} O_2(g) \quad (\Delta G^\circ = 92.34 \, \text{kJ mol}^{-1}) $ is $ 8.0 \times 10^{-3} $ at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation ($ \alpha $) of water is _____ $\times 10^{-2}$ (nearest integer value). [Assume $ \alpha $ is negligible with respect to 1]
One mole of an ideal gas at 300 K is compressed isothermally from a volume of \(V_1\) to \(V_2\). Calculate:
The work done on the gas
The change in internal energy
The heat exchanged with the surroundings
Use \(R = 8.314\, \text{J/molK}\), \( \ln(2.5) = 0.916\)