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put the following in order of increasing standard molar entropy. questi…

Question

put the following in order of increasing standard molar entropy.
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correct answer list
c2h6(l) hcooh(l) ch3cooh(l) ch4(l)

Explanation:

Response

To determine the order of increasing standard molar entropy, we use the following principles:

Step 1: Recall Entropy Trends

Standard molar entropy (\(S^\circ\)) depends on:

  • Molecular complexity: More atoms/branched structures → higher \(S^\circ\) (more microstates).
  • Phase: Gases > Liquids > Solids (but all here are liquids, so focus on molecular structure).
Step 2: Analyze Each Compound
  1. \(\boldsymbol{CH_4(l)}\) (Methane, liquid): Simplest structure (1 C, 4 H). Fewest atoms → lowest entropy.
  2. \(\boldsymbol{HCOOH(l)}\) (Formic acid): Structure: \( \text{HCOOH} \) (2 C, 2 O, 2 H). More complex than \(CH_4\) but less than larger molecules.
  3. \(\boldsymbol{CH_3COOH(l)}\) (Acetic acid): Structure: \( \text{CH}_3\text{COOH} \) (2 C, 2 O, 4 H). More atoms than \(HCOOH\) (longer carbon chain, more H).
  4. \(\boldsymbol{C_2H_6(l)}\) (Ethane, liquid): Structure: \( \text{C}_2\text{H}_6 \) (2 C, 6 H). Longer carbon chain than \(CH_4\), with more H atoms than \(HCOOH\) or \(CH_3COOH\) in terms of C - H bonds (simpler than carboxylic acids but more complex than \(CH_4\) or \(HCOOH\)? Wait, no—carboxylic acids have O atoms, introducing more complexity (double bonds, polar groups). Wait, correction:

Wait, re - evaluate:

  • \(CH_4(l)\): 1 C, 4 H (simplest).
  • \(HCOOH(l)\): 2 C, 2 O, 2 H (has C = O, O - H).
  • \(CH_3COOH(l)\): 2 C, 2 O, 4 H (longer alkyl group, more H).
  • \(C_2H_6(l)\): 2 C, 6 H (only C - C and C - H bonds, no O).

But entropy also depends on the number of atoms and molecular flexibility. Carboxylic acids have polar groups (O - H, C = O) which restrict rotation less than alkanes? No—alkanes have free rotation around C - C bonds. Wait, actually, the key is number of atoms and molecular weight (more atoms/weight → more microstates).

Let’s count total atoms:

  • \(CH_4\): \(1 + 4 = 5\) atoms.
  • \(HCOOH\): \(1 (H) + 1 (C) + 2 (O) + 1 (H) = 5\) atoms? Wait, no: \(HCOOH\) is \(H - C(=O) - O - H\), so atoms: H, C, O, O, H → 5 atoms? Wait, \(C_2H_6\) is \(C - C - H\) (with 6 H: \(H_3C - CH_3\)), so 2 C + 6 H = 8 atoms. \(CH_3COOH\) is \(H_3C - C(=O) - O - H\): 2 C, 2 O, 4 H → 8 atoms. \(HCOOH\) is \(H - C(=O) - O - H\): 1 C, 2 O, 2 H → 5 atoms.

Wait, I made a mistake earlier. Let's re - list:

  • \(CH_4(l)\): \(C + 4H = 5\) atoms.
  • \(HCOOH(l)\): \(C + 2O + 2H = 5\) atoms.
  • \(CH_3COOH(l)\): \(2C + 2O + 4H = 8\) atoms.
  • \(C_2H_6(l)\): \(2C + 6H = 8\) atoms.

Now, for molecules with the same number of atoms, complexity (functional groups, branching) matters. \(HCOOH\) has a C = O and O - H (polar, more complex than \(CH_4\)’s single bonds). So \(CH_4 < HCOOH\) (same atom count, but \(HCOOH\) is more complex).

For 8 - atom molecules: \(CH_3COOH\) has O atoms (C = O, O - H) and \(C_2H_6\) is a simple alkane. The polar groups in \(CH_3COOH\) create more microstates (due to dipole - dipole interactions and different bond types) than the non - polar \(C_2H_6\)? Wait, no—alkanes have free rotation around C - C bonds, but carboxylic acids have restricted rotation around C = O. Wait, actually, the correct trend for these liquids (all are liquid at standard conditions? Wait, no—\(CH_4\) is gas at STP, but the problem says “(l)”, so we assume liquid phase for all.

The correct order (from lowest to highest \(S^\circ\)) is:

  1. \(CH_4(l)\) (simplest, fewest atoms, least complex).
  2. \(HCOOH(l)\) (more complex than \(CH_4\), same atom count, polar groups).
  3. \(C_2H_6(l)\) (8 atoms, simple alkane, more atoms than first two).
  4. \(CH_3COOH(l)\) (8 atoms, carboxylic acid, more complex than \(C_2H_6\) due to O…

Answer:

\(CH_4(l)\), \(HCOOH(l)\), \(C_2H_6(l)\), \(CH_3COOH(l)\) (in order of increasing standard molar entropy)