To calculate the standard enthalpy change for the reaction of [Co H2O 6]2+ aq with 2 Cl- aq to form [CoCl4]2- aq and 6 H2O l , we can use Hess's Law. We can manipulate the given reactions to get the desired reaction and then add their enthalpy changes.The desired reaction is:[Co H2O 6]2+ aq + 2 Cl- aq [CoCl4]2- aq + 6 H2O l First, we can multiply the second given reaction by 2 to get the same amount of Cl- ions as in the desired reaction:2 [Co H2O 6]2+ aq + 2 Cl- aq [CoCl4]2- aq + 4 H2O l 2[Co H2O 6]2+ aq + 4 Cl- aq 2[CoCl4]2- aq + 8 H2O l H= -131 kJ mol-1Now, we can subtract the first given reaction from the modified second reaction: 2[Co H2O 6]2+ aq + 4 Cl- aq 2[CoCl4]2- aq + 8 H2O l - [Co H2O 6]2+ aq + 4 Cl- aq [CoCl4]2- aq + 6 H2O l This results in the desired reaction:[Co H2O 6]2+ aq + 2 Cl- aq [CoCl4]2- aq + 6 H2O l Now, we can find the enthalpy change for the desired reaction by subtracting the enthalpy change of the first given reaction from the modified second reaction:H desired = H modified second - H first H desired = -131 kJ mol-1 - -98.7 kJ mol-1 H desired = -32.3 kJ mol-1So, the standard enthalpy change for the reaction of [Co H2O 6]2+ aq with 2 Cl- aq to form [CoCl4]2- aq and 6 H2O l at a temperature of 298 K is -32.3 kJ mol-1.