consider the following
CO(g) + H2O(g)---> CO2(g) + H2(g) kc= 5.0
at equalibrium a 2.0 L container was found to contain 0.20 mol of CO2(g) , 0.30 mol of H2O(g) and 0.90 mol of H2(g).
The equalibirum amount of carbon dioxide is
question 2
4HCl(g) + O2(g) ---> 2H2O(g) + 2Cl2(g) kc= 357
An equalibirum mixture of the above system was found to contain the following concentratins: HCl(g)= 0.35mol/L, O2(g)= 0.078 mol/L and H2O(g)= 1.2mol/L. the equalibirum concentration of Cl2(g) is?
6 years ago
Answered By Leonardo F
This question is about the equilibrium constant and how to express it in terms of the concentrations of the chemical compounds involved in the reaction. For question 1, we can express the equilibrium contatnt (kc) as a fraction between the concentrations of the products and the concentration of the reactants, each powered by their own stoichiometric coefficient. Hence, we can write:
$kc=\frac{\left[CO_2\right]\left[H_2\right]}{\left[CO\right]\left[H_2O\right]}$kc=[CO2][H2][CO][H2O]
We know the number of moles of all the species, except carbon dioxide. The amount concentration can be calculated as the number of moles of that species divided by the volume of the solution. The only problem that I see is: the number 0.2 mol is for carbon dioxide or cabon monoxide? I ask this because if it is for carbon dioxide, there is no point at having all the other information about kc, because we could simply divide 0.2 mol / 2.0 L, to get the equilibrium concentration of CO2. I am going to consider that 0.2 mol is for CO, not for CO2. Hence:
$\left[CO\right]=\frac{0.20mol}{2.0L}$[CO]=0.20mol2.0L = 0.10 mol/L
$\left[H_2O\right]=\frac{0.30mol}{2.0L}$[H2O]=0.30mol2.0L = 0.15 mol/L
$\left[H_2\right]=\frac{0.90mol}{2.0L}$[H2]=0.90mol2.0L = 0.45 mol/L
Putting all of these values in the kc expression, we will have:
$kc=5.0=\frac{\left[CO_2\right]\left(0.45\right)}{\left(0.10\right)\left(0.15\right)}$kc=5.0=[CO2](0.45)(0.10)(0.15)
Isolating $\left[CO_2\right]$[CO2] in the above expression:
$\left[CO_2\right]=\frac{\left(5.0\right)\left(0.10\right)\left(0.15\right)}{0.45}$[CO2]=(5.0)(0.10)(0.15)0.45 = 0.17 mol/L
Hence, the equilibrium concentration of carbon dioxide is approximately 0.17 mol/L.
Question 2) We have to do exactly the same thing that we did for the previous question. Writing the expression for kc:
$kc=\frac{\left[H_2O\right]^2\left[Cl_2\right]^2}{\left[HCl\right]^4\left[O_2\right]}$kc=[H2O]2[Cl2]2[HCl]4[O2]
Putting the values informed by the question:
$357=\frac{\left(1.2\right)^2\left[Cl_2\right]^2}{\left(0.35\right)^4\left(0.078\right)}$357=(1.2)2[Cl2]2(0.35)4(0.078)
Isolating $\left[Cl_2\right]$[Cl2] :
$\left[Cl_2\right]=\sqrt{\frac{\left(357\right)\left(0.35^4\right)\left(0.078\right)}{\left(1.2^2\right)}}$[Cl2]=√(357)(0.354)(0.078)(1.22) = 0.53 mol/L
Hence, the equilibrium concentration of chlorine gas is approximately 0.53 mol/L.