To determine the correct conditions under which the forward reaction is favored for the equilibrium:
\(\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \quad \Delta H = -Q\)
Let's analyze the factors influencing this reaction:
Considering these points, the correct answer is high pressure, low temperature, and higher concentration of \(\text{H}_2\), which would drive the equilibrium position towards the production of more ammonia, as justified by Le Chatelier's principle.
Conclusion: The option high pressure, low temperature, and higher concentration of H2 correctly describes the conditions that favor the forward direction of this equilibrium reaction.
At a given temperature and pressure, the equilibrium constant values for the equilibria are given below:
$ 3A_2 + B_2 \rightleftharpoons 2A_3B, \, K_1 $
$ A_3B \rightleftharpoons \frac{3}{2}A_2 + \frac{1}{2}B_2, \, K_2 $
The relation between $ K_1 $ and $ K_2 $ is:
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: