Using given data, calculate the change in Gibbs free energy for each of the following reactions. You may want to reference (Pages 831 - 832) Section 19.6 while completing this problem. Part A: 2Ag(s)+Cl2(g)→2AgCl(s) Gibbs free energy for AgCl(s) is −109.70 kJ/mol Express your answer without scientific notation and using one decimal place.(units kJ) Part B: P4O10(s)+16H2(g)→4PH3(g)+10H2O(g) Gibbs free energy for P4O10(s) is −2675.2 kJ/mol Gibbs free energy for PH3(g) is 13.4 kJ/mol Gibbs free energy for H2O(g) is −228.57 kJ/mol Express your answer without scientific notation and using one decimal place.(units kJ) Part C: CH4(g)+4F2(g)→CF4(g)+4HF(g) Gibbs free energy for CH4(g) is −50.8 kJ/mol Gibbs free energy for CF4(g) is −635.1 kJ/mol Gibbs free energy for HF(g) is −270.70 kJ/mol Express your answer without scientific notation and using one decimal place. (unitskJ) 2H2O2(l)→2H2O(l)+O2(g) Gibbs free energy for H2O2(l) is −120.4 kJ/mol Gibbs free energy for H2O(l) is −237.13 kJ/mol Express your answer without scientific notation and using one decimal place.(units kJ) Part D: 2H2O2(l)→2H2O(l)+O2(g) Gibbs free energy for H2O2(l) is −120.4 kJ/mol Gibbs free energy for H2O(l) is −237.13 kJ/mol