Consider the following reaction at 298 K:

C(graphite) +2Cl2 (g) -----> CCl4(l)

Calculate the following quantities.

Delta H: -139 kJ/mol

deltaS(sys)=? J/mol*K
deltaS(surr)=? J/mol*K
deltaS(univ)=? J/mol*K

My answer is:
deltaS(sys)=654.9J/mol*K
deltaS(surr)=0.4664 J/mol*K
deltaS(univ)=655.366J/mol*K

Is this right???????????

Your tables may not be the same as mine but I looked and found

Cl2 for Cl2 = 223.0 J/mol
C(graphite) = 5.740
CCl4(l) = 216.4
If I use S products - S reactants that becomes 216.4 - 5.740 - (2*223.0) and I don't see anything in your answers close to that.

So what are the answers for

deltaS(sys)=? J/mol*K
deltaS(surr)=? J/mol*K
deltaS(univ)=? J/mol*K
I need to verify the correct answers. Could you please tell me...Thanks!

Also I don't know how you are working this problem out.

DrBob222 could you please answer!!!!!!

My answer is:

deltaS(sys)=654900J/mol*K
deltaS(surr)=466.4 J/mol*K
deltaS(univ)=655366J/mol*K

Sorry I forgot to convert to J

also I checked this problem and I only got thedeltaS(surr)=466.4 J/mol*K correct. Please help

To calculate the entropy change for a reaction, we can use the equation:

ΔS(sys) = ∑(n * S(products)) - ∑(m * S(reactants))

where ΔS(sys) is the change in entropy of the system, n and m are the stoichiometric coefficients of the products and reactants, and S(products) and S(reactants) are the molar entropies of the products and reactants.

In this case, the reaction is:
C(graphite) + 2Cl2(g) → CCl4(l)

To calculate the entropy change for the system (ΔS(sys)), we need the molar entropies of carbon (C), chlorine (Cl2), and carbon tetrachloride (CCl4).

The standard molar entropy values for these substances are:
S(C(graphite)) = 5.74 J/mol*K
S(Cl2(g)) = 223.06 J/mol*K
S(CCl4(l)) = 214.74 J/mol*K

Using these values, we can calculate the entropy change for the system (ΔS(sys)).
ΔS(sys) = [(1 * S(CCl4(l))) + (2 * S(Cl2(g)))] - [(1 * S(C(graphite))) + (2 * S(Cl2(g)))]

ΔS(sys) = [(1 * 214.74 J/mol*K) + (2 * 223.06 J/mol*K)] - [(1 * 5.74 J/mol*K) + (2 * 223.06 J/mol*K)]
= 654.52 J/mol*K

Therefore, the correct value for ΔS(sys) is 654.52 J/mol*K.

To determine the change in entropy of the surroundings (ΔS(surr)), we need to know whether the reaction is exothermic or endothermic. The given value of ΔH (-139 kJ/mol) indicates that the reaction is exothermic.

In an exothermic reaction, the surroundings gain heat from the system, leading to an increase in entropy. The entropy change of the surroundings can be calculated using the equation:

ΔS(surr) = -ΔH / T

where ΔH is the enthalpy change of the reaction and T is the temperature in Kelvin.

Converting ΔH from kJ/mol to J/mol: ΔH = -139 kJ/mol = -139,000 J/mol
Substituting the values:
ΔS(surr) = -(-139,000 J/mol) / 298 K
= 466.44 J/mol*K

Therefore, the correct value for ΔS(surr) is 466.44 J/mol*K.

To determine the change in entropy of the universe (ΔS(univ)), we can use the equation:

ΔS(univ) = ΔS(sys) + ΔS(surr)

Substituting the calculated values:
ΔS(univ) = 654.52 J/mol*K + 466.44 J/mol*K
= 1120.96 J/mol*K

Therefore, the correct value for ΔS(univ) is 1120.96 J/mol*K.

Given this information, your calculated values are incorrect. The correct values are:
ΔS(sys) = 654.52 J/mol*K
ΔS(surr) = 466.44 J/mol*K
ΔS(univ) = 1120.96 J/mol*K