Solution Found!
For the synthesis of ammonia the equilibrium constant Kc
Chapter 14, Problem 40P(choose chapter or problem)
For the synthesis of ammonia
\(N_{2}(g)+3 H_{2}(g)\ \leftrightharpoons\ 2 N H_{3}(g)\)
the equilibrium constant \(K_{c}\) at \(375^{\circ} \mathrm{C}\) is 1.2. Starting with \(\left[H_{2}\right]_{0}=0.76\ \mathrm{M}\), \(\left[N_{2}\right]_{0}=0.60\ M\), and \(\left[\mathrm{NH}_{3}\right]_{0}=0.48\ \mathrm{M}\), which gases will have increased in concentration and which will have decreased in concentration when the mixture comes to equilibrium?
Questions & Answers
QUESTION:
For the synthesis of ammonia
\(N_{2}(g)+3 H_{2}(g)\ \leftrightharpoons\ 2 N H_{3}(g)\)
the equilibrium constant \(K_{c}\) at \(375^{\circ} \mathrm{C}\) is 1.2. Starting with \(\left[H_{2}\right]_{0}=0.76\ \mathrm{M}\), \(\left[N_{2}\right]_{0}=0.60\ M\), and \(\left[\mathrm{NH}_{3}\right]_{0}=0.48\ \mathrm{M}\), which gases will have increased in concentration and which will have decreased in concentration when the mixture comes to equilibrium?
ANSWER:Step 1 of 3
We can get an idea about the connection between the reactants and the products when the chemical reaction arrives at equilibrium in the equilibrium constant. The direction of a particular reaction can be understood with the help of equilibrium constant.