Consider the following endothermic equilibrium: N_2O_4 ReverseEquilibrium 2NO_2(
ID: 992403 • Letter: C
Question
Consider the following endothermic equilibrium: N_2O_4 ReverseEquilibrium 2NO_2(g) The number of moles of N_2O_4 at equilibrium could be increased by adding a catalyst increasing the temperature increasing the volume by decreasing the pressure. adding NO_2 More than one answer is correct For the reaction shown below. K_eq=2 times 10^11. Which of the following statements concerning this system at equilibrium is true? 2CO(g) + O_2(g) ReverseEquilibrium 2CO_2(g) The equilibrium favors reactants. The equilibrium solution contains equal amounts of CO, O_2, and CO_2. The reaction is very fast, due to the high value for K_eq. The equilibrium solution contains predominantly CO_2. The equilibrium system contains almost twice as many reactant molecules as product molecules. As the [H_3O^+] in a solution increases, the [OH^-] increases and the pH increases. increases and the pH decreases. decreases und the pH increases. decreases and the pH decreases. The conjugate base of H_2PO_4^- is PO_4^3- HPO_4^- HPO_4^2- H_3PO_4 None of the above Consider the following reaction: HC_2O4^- + H_2BO_3^- ReverseEquilibrium H_3BO_3 + C_2O_4^2- Identify the acid and it's conjugate base in that order. H_2BO_3^- and H_3BO_3 HC_2O_4^- and H_3BO_3 HC_2O_4^- and C_2O_4^2- H_2BO_3^- and C_2O_4^2-Explanation / Answer
8.
1 mol of reactant, 2 mol of product
so:
adding a catalys has no effect in equilibrium, only in rate
increasing T will favour reaction, since this is endothermic, meaning it needs energy, it won't favour N2O4 formation
increasing V, decreasing P, will favour the most mol formation, this will not favour N2O4
adding NO2 will shift toward reactants; which is N2O4, choose thi one
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