Calculate the standard enthalpy of formation of solid Mg(OH)_2, given the follow

Question

Calculate the standard enthalpy of formation of solid Mg(OH)_2, given the following data: 2 Mg(s) + O_2(g) rightarrow 2MgO(s) Delta H degree = -1203.6kJ Mg(OH)_2 rightarrow MgO(s) + H_2O(l) Delta H degree = +37.1 kJ 2 H_2(g) + O_2 rightarrow 2 H_2O(l) Delta H degree = -571.7 kJ Calculate the standard enthalpy of formation of gaseous diborane (B_2H_6) using the following thermochemical information: 4 B(s) + 3 O_2 rightarrow 2 B_2O_3(s) Delta H degree = -2509.1 kJ 2 H_2(g) + O_2(g) rightarrow 2 H_2O(l) Delta H degree = -571.7 kJ

Explanation / Answer

Ans. Dividing eqn 1 by 2

2Mg(s) + O2(g) -----> 2MgO(s) Delta H = -1203.6kJ

Dividing by 2,

Mg(s) + 1/2O2(g)------> MgO(s) Delta H = -601.8kJ .....(1)

Eqn 3 is:

2H2(g) + O2(g) --------> 2H2O(l) Delta H = -571.71kJ

Dividing eqn by 2

H2(g) + O2(g) ------> H2O(l) Delta H = -285.85kJ......(2)

Eqn 2 is

Mg(OH)2(s)------> MgO(s) + H2O(l) Delta H = 37.1kJ

Reversing the equation:

MgO(s) + H2O(l) ------> Mg(OH)2 (s) Delta H = -37.1kJ....(3)

Thus heat of formation of Mg(OH)2 is the sum of delta H from 1,2 and 3 = -601.8+(-285.85) +(-37.1) = -924.75kJ

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