Question
For the reaction 2NH3(g) 3H2(g) N2(g) at 472oC equilibrium is established when [H2] =0.0200M, [N2] = 0.0100M and [NH3] = 5.37x10^_8 M. What is the equilibrium constant for the reverse reaction?
Answers
Write the equilibrium constant expression and substitute the numbers from the problem. Then solve for Kc
Related Questions
For the reaction mc011-1.jpg at 472°C equilibrium is established when [H2] =0.0200M, [N2] = 0.0100M...
For the reaction: N2+3H2 <-----> 2NH3
1.000 mol N2 and 1.000 mol H2 are placed in a 1.000 L contain...
For the reaction: N2+3H2<--->2NH3
4.000 mol N2, 2.000 mol H2, and 6.000 mol NH3 are placed in a 2.0...