#ΔG_text(rxn) = "-49.2 kJ·mol"^"-1"#. The reaction is spontaneous under these conditions.
The equation for the reaction is
#"ATP(aq) + H"_2"O(l)" ⇌ "ADP(aq) + HPO"_4^"2-"#
The equation for free energy #ΔG# is
#ΔG = ΔG^@ + RTlnQ#
where
#ΔG^@ =# the standard free energy change for the reaction
#R =# the Ideal Gas Constant
#T=# the Kelvin temperature
#Q = (["ADP"]["HPO"_4^"2-"])/(["ATP"]#
In this problem,
#["ADP"] = 0.70 × 10^"-3" color(white)(l)"mol/L"#
#["HPO"_4^"2-"] = "5.0 mmol/L"#
#["ATP"] = "5.0 mmol/L"#
#R = "8.314 J·K"^"-1""mol"^"-1"#
#T = "(37.0 + 273.15) K = 310.15 K"#
∴ #Q= (0.70 × 10^"-3" × color(red)(cancel(color(black)(5.0))))/color(red)(cancel(color(black)(5.0))) = 7.0 × 10^"-4"#
#ΔG = "-30 500 J·mol"^"-1" + "8.314 J"·color(red)(cancel(color(black)("K"^"-1")))"mol"^"-1" × 310.15 color(red)(cancel(color(black)("K"))) × ln(7.0 × 10^"-4") = "30 500 J·mol"^"-1" -
"18 700 J·mol"^"-1" = "-49 200 J·mol"^"-1" = "-49.2 kJ·mol"^"-1"#
