Given: 2N2(g) +O(2) --> 2N2O(g)

2N2O+O2=4NO calculate enthalpy
1. 1 answer
2. asked by Yogesh
3. 693 views
consider the reaction 2N2O(g) == O2(g)+2N2(g).which of the following will cause a shift in the equilibrium to the left?
1. 1 answer
2. asked by sara
3. 2,189 views
Given Kc values:N2(g)+1/2 O2(g)<->N2O(g) Kc=2.7* 10^-18 N2O4(g)<->2NO2(g) Kc=4.6*10^-3 1/2N2(g)+O2(g)<->NO2(g) kc=4.1*10^-9 What
1. 3 answers
2. asked by Juliet
3. 2,217 views
Using Hess’s Law with the following data:N2(g) + O2(g) = 2NO(g) Del(H) = 180.7kj 2NO(g) + O2(g) = 2NO2(g) Del(H)= -113.1kj
1. 1 answer
2. asked by Dan
3. 1,198 views
2N2O(g) → 2N2(g) + O2(g)d[N2O]/dt = -5.8×10-4 M/s at a particular temperature and set of concentrations. What are d[O2]/dt
1. 0 answers
2. asked by Jane
3. 668 views
2N2O(g) → 2N2(g) + O2(g)d[N2O]/dt = -5.8×10-4 M/s at a particular temperature and set of concentrations. What are d[O2]/dt
1. 0 answers
2. asked by Jane
3. 577 views
Given the equilibrium constant values:1.N2(g)+ 1/2O2(g)<---> N2O(g) KC=2.7*10^{-18} 2.N2O4(g)<----> 2NO2(g) KC= 4.6*10^{-3} 3.
1. 3 answers
2. asked by Saira
3. 1,448 views
I have a question about this one2N2O(g) ⇋ O2(g) + 2N2(g), what happens to the equilibrium position if the pressure decreases?
1. 4 answers
2. asked by alex
3. 6,169 views
Given the equilibrium constant values:N2 + 1/2O2 >>> N2O ; kc = 2.7 * 10^-18 N2O4>>>>2NO2 kc=4.6*10^-3 1/2N2 + O2>>>NO2 kc = 4.1
1. 3 answers
2. asked by abe
3. 2,161 views
Hey guys! Stuck on another one: The reaction 2N2O(g) 2N2(g) + O2(g) has Kc = 3.5 × 10-18 at a particular temperature. If 0.20
1. 2 answers
2. asked by Allie
3. 953 views