Which reaction below would represent the standard enthalpy of formation for CHCl3 at 25°C? Hint: The reaction must be balanced. Fractional coefficients are ok in heats of formation. All reactants must be elements in their standard state (solid, liquid, or gas) at 25°C. Remember that certain elements exist as diatomics (O2, N2, F2, H2, I2, Cl2, Br2).Group of answer choicesCHCl3(l) → C(s) + H(g) + 3 Cl(g)C(s) + H(g) + 3 Cl(g) → CHCl3(l)C(s) + 1/2 H2(g) + 3/2 Cl2(g) → CHCl3(l)2 C(s) + H2(g) + 3 Cl2(g) → 2 CHCl3(l)
Question
Which reaction below would represent the standard enthalpy of formation for CHCl3 at 25°C? Hint: The reaction must be balanced. Fractional coefficients are ok in heats of formation. All reactants must be elements in their standard state (solid, liquid, or gas) at 25°C. Remember that certain elements exist as diatomics (O2, N2, F2, H2, I2, Cl2, Br2).Group of answer choicesCHCl3(l) → C(s) + H(g) + 3 Cl(g)C(s) + H(g) + 3 Cl(g) → CHCl3(l)C(s) + 1/2 H2(g) + 3/2 Cl2(g) → CHCl3(l)2 C(s) + H2(g) + 3 Cl2(g) → 2 CHCl3(l)
Solution
The correct reaction that represents the standard enthalpy of formation for CHCl3 at 25°C is:
C(s) + 1/2 H2(g) + 3/2 Cl2(g) → CHCl3(l)
This is because the standard enthalpy of formation is defined as the change in enthalpy when one mole of a substance in the standard state (1 atm of pressure and 298.15 K) is formed from its pure elements under the same conditions. Here, Carbon (C) is in its standard state as a solid, Hydrogen (H2) and Chlorine (Cl2) are in their standard states as diatomic gases. They combine to form one mole of CHCl3 in its standard state as a liquid.
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To find the enthalpy of a reaction using heats of formation…
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