Fe 3+ + SCN- --> FeSCN 2+
<span>.......Fe 3+.......SCN-.........FeSCN 2+ </span>
<span>I.......0.04..........0.001.............. </span>
<span>C........-x...............-x............. </span>
<span>E.....0.04-x.....0.001-x...........x </span>
<span>Keq = 203.4 = x / (0.04-x)(0.001-x) </span>
<span>203.4 = x / (x^2 - 0.041x + 4x10^-5) </span>
<span>203.4x^2 - 8.34x + 0.00094 = x </span>
<span>203.4x^2 - 9.34x + 0.00094 = 0 </span>
<span>x = -0.0001M or 0.0458M </span>
<span>therefore, according to the calculated Keq, all of the SCN- and Fe 3+ would be fully converted into FeSCN 2+</span>
Answer:
The enthalpy of the second intermediate equation is altered by halving its value and changing the sign.
Explanation:
Let's examine both the first and second intermediate reactions alongside the overall equation concerning the examined process;
First reaction;
Ca (s) + CO₂ (g) + ½O₂ (g) → CaCO₃ (s) ΔH₁ = -812.8 kJ
Second reaction;
2Ca (s) + O₂ (g) → 2CaO (s) ΔH₂ = -1269 kJ
Thus, the overall reaction becomes;
CaO (s) + CO₂ (g) → CaCO₃ (s) ΔH =?
According to Hess's law, which states that the total heat change in a reaction is equal to the sum of the heat changes for each step, we cannot simply sum the enthalpies for this overall reaction. Instead, we obtain the overall enthalpy by halving the second intermediate reaction's enthalpy and changing its sign before adding, as illustrated below;
Enthalpy of Intermediate reaction 1 + ½(-Enthalpy of Intermediate reaction 2) = Enthalpy of Overall reaction
The pH level is 1.39. To explain, we start with the given information: the concentration of HClO is 0.15 M, with an acid dissociation constant of 2.9 × 10-8. The objective is to calculate the pH of the solution. Through the process, we find that the equilibrium concentration after applying the formula yields 0.04069 M for H3O⁺, leading us to a pH of 1.39.
