Question

What is the value of $\Delta {S_{surr}}$ for the following reaction at $298K$-
$6C{O_2}(g) + 6{H_2}O(g) \to {C_6}{H_{12}}{O_6}(g) + 6{O_2}(g)$
Given that: $\Delta {G^ \circ } = 2879 KJ{\text{ mo}}{{\text{l}}^{ - 1}}$ $\Delta S = - 210J{\text{ }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}$

Answer 1

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

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.