Draw enantiomers for the following compounds using

Perspective Formulas
Fischer Projections

The reaction shown here has a K_p=4.5*10² at 825K. Find Kc for the reaction at this temperature:

CH₄(g) + CO₂(g) <—> 2CO(g) + 2H₂(g)

The Correct Answer and Explanation is:

Part 1: Drawing Enantiomers

Let’s assume the compound of interest is lactic acid (CH₃CH(OH)COOH) — a common example with one chiral center. If your compound is different, please specify.

a) Perspective Formulas (3D Wedge-Dash)

Lactic acid has one chiral center at the second carbon (C2).

(R)-Lactic Acid:

         OH
          |
H3C — C — COOH
          |
         H

Perspective (Wedge-Dash):

       OH
        |
H3C — C — COOH
     /   \
   H     (wedge) OH
          (dash) H

(S)-Lactic Acid:
Swap two groups to get the enantiomer:

       OH
        |
H3C — C — COOH
     /   \
(wedge) H   (dash) OH

b) Fischer Projections

(R)-Lactic Acid:

      COOH
       |
H — C — OH
       |
      CH3

(S)-Lactic Acid:

      COOH
       |
OH — C — H
       |
      CH3

Part 2: Finding KcK_c from KpK_p

Given:

Kp=4.5×102K_p = 4.5 \times 10^2
Temperature, T=825T = 825 K
Reaction: CH4(g)+CO2(g)⇌2CO(g)+2H2(g)\text{CH}_4(g) + \text{CO}_2(g) \rightleftharpoons 2\text{CO}(g) + 2\text{H}_2(g)

Step 1: Use the equation to relate KpK_p and KcK_c: Kp=Kc(RT)ΔnK_p = K_c(RT)^{\Delta n}

Where:

R=0.0821 L\cdotpatm/mol\cdotpKR = 0.0821 \, \text{L·atm/mol·K}
Δn=moles of gas products−moles of gas reactants=(2+2)−(1+1)=2\Delta n = \text{moles of gas products} – \text{moles of gas reactants} = (2+2) – (1+1) = 2

Step 2: Solve for KcK_c: Kc=Kp(RT)Δn=4.5×102(0.0821×825)2K_c = \frac{K_p}{(RT)^{\Delta n}} = \frac{4.5 \times 10^2}{(0.0821 \times 825)^2}

Calculate: RT=0.0821×825=67.7325RT = 0.0821 \times 825 = 67.7325 (RT)2=(67.7325)2≈4586.73(RT)^2 = (67.7325)^2 \approx 4586.73 Kc=4504586.73≈0.0981K_c = \frac{450}{4586.73} \approx 0.0981

Final Answer:

Kc≈0.0981\boxed{K_c \approx 0.0981}

Explanation

In gas-phase equilibrium reactions, the equilibrium constants KpK_p and KcK_c are related by the formula: Kp=Kc(RT)ΔnK_p = K_c(RT)^{\Delta n}

where RR is the gas constant, TT is the absolute temperature in Kelvin, and Δn\Delta n is the change in the number of moles of gas between products and reactants. This relationship accounts for how pressure and concentration differ in behavior for gases.

In this case, the reaction involves 2 moles of CO and 2 moles of H₂ (total 4) on the product side, and 1 mole of CH₄ and 1 mole of CO₂ (total 2) on the reactant side. So Δn=4−2=2\Delta n = 4 – 2 = 2.

We are given Kp=4.5×102K_p = 4.5 \times 10^2 at 825 K. Using the ideal gas constant R=0.0821 L\cdotpatm/mol\cdotpKR = 0.0821 \, \text{L·atm/mol·K}, we first compute RTRT, square it (since Δn=2\Delta n = 2), and then solve for KcK_c by dividing KpK_p by this quantity.

The calculation gives: Kc=450(0.0821×825)2=4504586.73≈0.0981K_c = \frac{450}{(0.0821 \times 825)^2} = \frac{450}{4586.73} \approx 0.0981

This result means that even though the partial pressures strongly favor the products (large KpK_p), the actual concentration-based constant KcK_c is lower due to the pressure-concentration conversion that accounts for the increase in gas moles (which dilutes concentration but increases pressure).