There is no coefficient in front of the $N_2$ in the reactants. This means that there is one mole of $N_2$ required in the balanced equation. There is a...
1 Answers 1 viewsThree equiv of hydrogen react with 1 equiv of nitrogen to give one equiv ammonia. Of course, I could double this reaction to remove the 1/2 coefficients, but the will...
1 Answers 1 viewsAccording to Dalton's law of partial pressures, $P_(t otal) = P_1 + P_2 + P_3 ...$ So the total pressure of the system is merely the partial pressures of each...
1 Answers 1 viewsWe always have to read the problems completely to understand the question. In this case even when we have pressure and temperature conditions specified, they don't matter, first because they...
1 Answers 1 viewsBefore doing any calculation, take a look at the values of the equilibrium concentrations for the three chemical species that take part in the reaction. Notice that the concentrations of...
1 Answers 1 viewsAs you know, the equilibrium constant for a given chemical equilibrium depends on the equilibrium concentrations of the chemical species that take part in the reaction the stoichiometric...
1 Answers 1 viewsUse the definition of $K_(eq)$. $K_(eq) = (["products"]^(d,e,f, . . . ))/(["reactants"]^(a,b,c,. . . ))$ $= (["NH"_3]_(eq)^2)/(["H"_2]_(eq)^3["N"_2]_(eq))$ Just make sure you remember to use stoichiometric coefficients correctly. $3H_2$...
1 Answers 1 viewsRemember to include the coefficients in the change in concentratio, as well as the exponents. I get $K_c = 1.69$. If $K_p = K_c(RT)^(Deltan_"gas")$, what would $K_p$ be if this...
1 Answers 1 views$HCl(aq)+H_2O(l)rarr Cu^(2+) + 2NO_3^-$ $Ca(OH)_2(s) stackrel(H_2O)rarr Ca^(2+) +2HO^-$ $Ca(NO_3)_2(s) stackrel(H_2O)rarr Ca^(2+) +2NO_3^-$ $NH_3(aq) +H_2O(l) rightleftharpoonsNH_4^+ + HO^-$ And one of these things is not like the other ones.........
1 Answers 1 viewsAmmonia is a Bronsted base: $NH_3(aq) + H_2O(l) rightleftharpoons NH_4^+ + OH^-$ $pK_b = 4.75$ I am not going to solve this equation using $pK_b$, but most of the ammonia...
1 Answers 1 views