Please help me
N2O5(g) + H2O(l) --> 2HNO3(l) delta H° = -76.2 kJ
H2O(l) --> H2(g) + 1/2O2(g) delta H° = 286.0 kJ
1/2N2(g) + 3/2O2(g) + 1/2H2(g) --> HNO3(l) delta H° = -174.0 kJ
Calculate delta H° for the reaction
2N2O5(g) --> 2N2(g) + 5O2(g)
Multiply equation 1 by 2 and add to the reverse of twice equation 2 and add 4 times the reverse of equation 3.
To calculate the delta H° for the reaction 2N2O5(g) --> 2N2(g) + 5O2(g), you can use the concept of Hess's Law.
Hess's Law states that the enthalpy change for a reaction is independent of the pathway taken, as long as the initial and final conditions are the same. Therefore, you can add or subtract known enthalpy changes of reactions to obtain the desired reaction.
In this case, you can use the given reactions and their enthalpy changes to calculate the delta H° for the desired reaction.
Step 1: Reverse the second reaction and adjust the coefficients:
H2(g) + 1/2O2(g) --> H2O(l)
(Delta H° = -286.0 kJ)
Step 2: Multiply the second reaction by 2 to match the coefficient of H2O in the first reaction:
2H2(g) + O2(g) --> 2H2O(l)
(Delta H° = 2*(-286.0 kJ) = -572.0 kJ)
Step 3: Multiply the second reaction by 1/2 to match the coefficient of O2 in the desired reaction:
1/2(2H2(g) + O2(g) --> 2H2O(l))
(H2(g) + 1/2O2(g) --> H2O(l))
(Delta H° = 1/2 *(-572.0 kJ) = -286.0 kJ)
Step 4: Multiply the third reaction by 2 to match the coefficient of HNO3 in the desired reaction:
2(1/2N2(g) + 3/2O2(g) + 1/2H2(g) --> HNO3(l))
(N2(g) + 3O2(g) + H2(g) --> 2HNO3(l))
(Delta H° = 2*(-174.0 kJ) = -348.0 kJ)
Step 5: Multiply the first reaction by 2 to match the coefficient of N2O5 in the desired reaction:
2(N2O5(g) + H2O(l) --> 2HNO3(l))
(2N2O5(g) + 2H2O(l) --> 4HNO3(l))
(Delta H° = 2*(-76.2 kJ) = -152.4 kJ)
Step 6: Add up all the modified reactions:
-152.4 kJ + (-286.0 kJ) + (-348.0 kJ) = -786.4 kJ
Therefore, the delta H° for the reaction 2N2O5(g) --> 2N2(g) + 5O2(g) is -786.4 kJ.