Question

A scientist measures the standard enthalpy change for the following reaction to be -146.9 kJ: (NH_4NO_3(aq) → N_2O(g) + 2 H_2O(l)). Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of (N_2O(g)) is:

A. -146.9 kJ mol^(-1)
B. 146.9 kJ mol^(-1)
C. 73.45 kJ mol^(-1)
D. -73.45 kJ mol^(-1)

Answer 1

The standard enthalpy of formation of N2O(g) is 30.6 kJ mol^(-1).

Step-by-step explanation:

The standard enthalpy of formation of N2O(g) can be calculated using the given information and the standard enthalpies of formation for the other substances involved in the reaction. The reaction is:

NH4NO3(aq) → N2O(g) + 2H2O(l)

Given that the standard enthalpy change for the reaction is -146.9 kJ and the standard enthalpy of formation for NH4NO3 is -177.5 kJ, we can calculate the standard enthalpy of formation for N2O(g) as follows:

Standard enthalpy change = sum of standard enthalpies of formation of products - sum of standard enthalpies of formation of reactants

-146.9 kJ = x - (-177.5 kJ)

x = -146.9 kJ - (-177.5 kJ)

x = -146.9 kJ + 177.5 kJ

x = 30.6 kJ

Therefore, the standard enthalpy of formation of N2O(g) is 30.6 kJ mol-1