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Question: What is the value of \(\Delta {S_{surr}}\) for the following reaction at \(298K\)- \(6C{O_2}(g) + ...

What is the value of ΔSsurr\Delta {S_{surr}} for the following reaction at 298K298K-
6CO2(g)+6H2O(g)→C6H12O6(g)+6O2(g)6C{O_2}(g) + 6{H_2}O(g) \to {C_6}{H_{12}}{O_6}(g) + 6{O_2}(g)
Given that: ΔG∘=2879KJ mol−1\Delta {G^ \circ } = 2879 KJ{\text{ mo}}{{\text{l}}^{ - 1}} ΔS=−210J K−1 mol−1\Delta S = - 210J{\text{ }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}

Explanation

Solution

To solve the given question, we have to use 2 equations. First will be the gibbs free energy equation, which would give us the value of enthalpy at the given temperature. And next, we will use that calculated enthalpy in the formula of ΔSsurr\Delta {S_{surr}} to find the value of ΔSsurr\Delta {S_{surr}}. Also, make sure that the units are the same throughout the calculation.

Complete answer:

  1. first we need to use the gibbs free energy equation i.e. ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S, where
    ΔG=\Delta G = gibbs free energy
    ΔH\Delta H= enthalpy of reaction
    T=T = temperature
    ΔS=\Delta S = entropy of system of the reaction
  2. substituting the given values from the questions in the gibbs free energy equation, we get:
    2879KJ mol−1=ΔH−(298×(−210J K−1 mol−1))2879KJ{\text{ mo}}{{\text{l}}^{ - 1}} = \Delta H - (298 \times ( - 210J{\text{ }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}))
    Converting kilojoule into joule for the gibbs free energy, we get:
    2879000=ΔH+2879000 = \Delta H + 62580 J mol−162580{\text{ J mo}}{{\text{l}}^{ - 1}}
    This gives: ΔH=225320 J mol−1\Delta H = 225320{\text{ J mo}}{{\text{l}}^{ - 1}}
  3. now, with the value of ΔH\Delta H and the given temperature, i.e. 298K298K , we can employ these two data into the formula, i.e., ΔSsurr=−ΔHT\Delta {S_{surr}} = - \dfrac{{\Delta H}}{T}
    Thus, substituting the value of enthalpy of the reaction and at the given temperature of 298K298K, we can find the value of ΔSsurr\Delta {S_{surr}}.
  4. The value of ΔSsurr\Delta {S_{surr}} is given is ΔSsurr=−ΔHT\Delta {S_{surr}} = - \dfrac{{\Delta H}}{T}
    ⇒ΔSsurr=−225320298\Rightarrow \Delta {S_{surr}} = - \dfrac{{225320}}{{298}}
    ⇒ΔSsurr=756.107 J mol−1\Rightarrow \Delta {S_{surr}} = 756.107{\text{ J mo}}{{\text{l}}^{ - 1}}
    Thus, the value of ΔSsurr\Delta {S_{surr}}= 756.107 J mol−1756.107{\text{ J mo}}{{\text{l}}^{ - 1}}

Note:
Always remember to do the basic conversions such as kilojoules into joules, Celsius into kelvin, grams into moles or kilograms, etc. The units while solving the question must be the same throughout and there should not be any error while doing the arithmetic portion. Also, stay cautious of the signs in the formula.