Use Hess\'s law to calculate G Solution Notice how there is only one NH3 in the

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Notice how there is only one NH3 in the target equation and it's on the right-hand side. That means we have to flip our first equation and divide it by two. Like this: (1/2)N2(g) + (3/2)H2(g) ---> NH3(g) ?H° = -46 kJ Notice that the sign changed on the ?H and its numerical value was cut in half. 2) The target equation has one NO2 and it's on the left-hand side, so we need to flip the second equation. Like this: NO2(g) + 2H2(g) ---> (1/2) N2(g) + 2H2O(l) ?H° = -170 kJ Notice that the sign changed on the ?H. 3) The ?H for the target equation is: -46 + -170 = -216 kJ Problem #2: Calculate the enthapy for the following reaction: N2(g) + 2O2(g) ---> 2NO2(g) ?H° = ??? kJ Using the following two equations: N2(g) + O2(g) ---> 2NO(g) ?H° = +180 kJ 2NO2(g) ---> 2NO(g) + O2(g) ?H° = +112 kJ Think about it and then on to the answer. Problem #3: Given the following data: 2NO(g) ---> N2(g) + O2(g) ?H = -180.6 kJ N2(g) + O2(g) + Cl2(g) ---> 2NOCl(g) ?H = +103.4 kJ Find the ?H of the following reaction: 2NOCl(g) ---> 2NO(g) + Cl2(g) Solution: 1) Flip first reaction, flip second reaction: N2(g) + O2(g) ---> 2NO(g) ?H = +180.6 kJ 2NOCl(g) ---> N2(g) + O2(g) + Cl2(g) ?H = -103.4 kJ 2) Add the equations and the ?H values: +180.6 + (-103.4) = +77.2 2NOCl(g) ---> 2NO(g) + Cl2(g) ?H = +77.2

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