? Part B g Poicy ants I Berfodc Tabl Calculate the enthalpy of the reaction 4B(s

Question

? Part B g Poicy ants I Berfodc Tabl Calculate the enthalpy of the reaction 4B(s) + 3O, (g)->2B2O3 (s) at released or ss is the same ace in one or in given the following perlinent information ??? = +36 kJ AH-285 kJ owing rdes 2 2B(s) + 3H2 (g)?B,14(g), 3H2(g)+3 02(g)?H2O(1), e added, their ded ersed, the sign anges of a reactiorn or, the enthalpy hat same factor Express your answer with the appeopriate units View Available Hints 5444 J Submit incorrect Try Again: 5 attempts remaining Provide Feedback MacBook Pro 8 UCPAL

Explanation / Answer

First we need to 2B2O3 in the right hand side, so multiply the reaction by 2 and reverse it, so the first required reaction will be

2B2H6(g) + 6O2(g) ------- 2B2O3(s) + 6H2O(g), Delta H1 = 2 * (-2035) = -4070 kJ

Now for getting 4B(s) we need to multiply the second reaction by 2

4B(s) + 6H2(g) --------- 2B2H6(g), Delta H2 = 2 * 36 = 72 kJ

adding both the reaction we get

4B(s) + 6H2(g) + 6O2(g) ------- 2B2O3(s) + 6H2O(g), Delta H3 = Delta H1 + Delta H2 = -3998 kJ

Now for cancelling O2 and H2, we need to 6 times the reaction 3 in reverse direction

6H2O(l) --------- 6H2(g) + 3O2(g), Delta H4 = 3 * 285 = 855 kJ

adding both the reactions, Delta H5 = Delta H3 + Delta H4 = -3143 kJ

4B(s) + 3O2(g) + 6H2O(l) ------- 2B2O3(s) + 6H2O(g) , Delta H5 = -3143 kJ

Now reversing the last H2O(l) ----- H2O(g) and adding it 6 times will yield the final reaction

6H2O(g) -------- 6H2O(l), Delta H6 = 6 * (-44) = -264 kJ

Hence final answer is -(3143+264) = -3407 kJ

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