The standard change in Gibbs free energy is ΔG°′=7.53 kJ/mol . Calculate ΔG for this reaction at 298 K when [dihydroxyacetone phosphate]=0.100 M and [glyceraldehyde-3-phosphate]=0.0010 0 M .

ΔG = -3879.6 J/mol = -3.88 kJ/mol

Explanation:

Step 1: Data given

The standard change in Gibbs free energy is ΔG°′=7.53 kJ/mol

Temperature = 298 K

[dihydroxyacetone phosphate]=0.100 M

[glyceraldehyde-3-phosphate]=0.00100 M .

Step 2: Calculate ΔG for this reaction

ΔG = ΔG° + RT ln ([glyceraldehyde-3-phosphate]/ [dihydroxyacetone phosphate])

⇒with ΔG° = 7.53 kJ/mol = 7

⇒with R = 8.314 J/mol*K

⇒with T = 298 K

⇒ with [glyceraldehyde-3-phosphate]=0.00100 M

⇒ with [dihydroxyacetone phosphate]=0.100 M

ΔG = 7530 J/mol + 8.314 * 298 * ln(0.001/0.1)

ΔG = -3879.6 J/mol = -3.88 kJ/mol

a). in this case, dh> 0 and sd> 0. if t becomes larger, the term "-tds" will become more negative. at some point, it is possible that -dts will be more negative than the dh is positive, and the reaction will become spontaneous. this is probably the correct answer, but you should look at all of the other possibilities and make sure of that.