The entropy change for a certain chemical reaction is positive. Is this enough i

Question

The entropy change for a certain chemical reaction is positive. Is this enough information to predict that this reaction is spontaneous? Explain The Delta G is calculated for a reaction at a given point in time and the value was a negative number. Is the reaction at equilibrium? Explain Which substance will have higher entropy Br_2 (s) or Br_2 (g) at the same temperature Ne (g) at 298 K or Ne (g) at 450 K CH_4 (g) or CH_4 (1) at 350 K Predict if Delta S^0_RXN will be , or = 0 for the examples below C_2H_2(g) + 5/2 O_2(g) rightarrow 2 CO_2(g) + H_2O(l) C(graphite) + O_2 (g) rightarrow CO_2 (g) C_2H_6 (g) + H_2O (g) rightarrow C_2H_5OH (l) + H_2 (g) Calculate Delta S^0_RXN for the reactions in question 4 Calculate Delta G^0_RXN for: CH_4(g) + 2O_2(g) rightarrow CO_2 (g) + 2 H_2O(g) Calculate the value of Delta G^0_RXN, at 298 K, using Delta H^0_RXN and Delta S^0_RXN for the reaction below C_2H_6 (g) + H_2O (g) rightarrow C_2H_5OH (l) + H_2 (g)

Explanation / Answer

1. Given the entropy change for a system is +ve, the information is not enough to find If the reaction is sponatenous or not.

2. When dG is -ve for a reaction, it is spontaneous reaction. dG is zero at equilibrium.

3. Higher entropy would be of,

a) Br2(g)

b) Ne(g) at 450 K

c) CH4(g)

4. dSo for the reaction,

a. dSo < 0

b. dSo > 0

c. dSo < 0

5. dSorxn

a) dSorxn = (2x 213.8 + 188.7) - (2.5 x 205 + 200.8) = -97 J/K.mol

b) dSorxn = (213.8) - (205 + 5.9) = 2.9 J/K.mol

c) dSorxn = (161.04 + 130.5) - (229.1 + 188.7) = -126.26 J/K.mol

6. dGo = (-137.2 - 228.4) - (-50.8) = -314.8 kJ/mol

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