Ask a New Question
Menu

Given:

Cu2O(s) + 1/2 O2(g) → 2CuO(s) deltaH°= -144 kJ
2Cu2O(s) → 2Cu(s) + 2CuO(s) deltaH°= 22 kJ

Calculate the standard enthalpy of formation of CuO(s).

1 answer

multiply 1 by 2 and add to the reverse of 2. That will give you 2Cu + O2 ==> 2CuO which is just twice what you want so divide by 2 for the equation and divide the dH by 2. Remember to multiply dH by 2 when multiplying the equation and remember to reverse the sign of dH when reversing the equation.
Similar Questions
  1. equilibrium constants for the following reactions4Cu(s) + O2(g) 2Cu2O(s), K1 2CuO(s) Cu2O(s) + 1/2O2(g), K2 what is K for the
    1. answers icon 1 answer
  2. Determine the missing values of deltaH in the diagram shown below^ | ^ N2O3(g) + 1/2 O2(g) (Enthalpy)| | ^ | | deltaH= +16.02 |
    1. answers icon 9 answers
  3. Which of the following is a correctly balanced equation?a. 2Cu + O2 → 2CuO b. 2Cu + 2O2 → 2CuO c. Cu + O2 → 2CuO d. 4Cu +
    1. answers icon 1 answer
  4. Given the following equations:2 H2 (g) + O2 (g)--> 2 H2O (l) deltaH = -571.6 kJ N2 (g) + O2 (g)-->2 NO (g) deltaH = +180.5 kJ N2
    1. answers icon 1 answer
more similar questions