10) When 1.50 g of Ba( s ) is added to 100.00 g of water in a container open to

Question

10) When 1.50 g of Ba( s ) is added to 100.00 g of water in a container open to the atmosphere, the reaction shown below occurs and the temperature of the resulting solution rises from 22.00 ° C to 33.10 ° C. If the specific heat of the solution is 4.18 J/(g · ° C), calculate H for the reaction, as written. Ba( s ) + 2 H 2 O( l ) Ba(OH) 2 ( aq ) + H 2 ( g ) H = ? 10) 11) For the reaction 2CH 4 ( g ) + 3 Cl 2 ( g ) 2 CHCl 3 ( l ) + 3 H 2 ( g ), H ° = - 118.6 kJ. H ° f = - 134.1 kJ/mol for CHCl 3 ( l ). Find H ° f for CH 4 ( g ). 11) 12) 2 LiOH(s) Li 2 O(s) + H 2 O(l) H ° = 379.1 kJ LiH(s) + H 2 O(l) LiOH(s) + H 2 (g) H ° = - 111.0 kJ 2 H 2 (g) + O 2 (g) 2 H 2 O(l) H ° = - 285.9 kJ Compute H ° in kJ for 2 LiH(s) + O 2 (g) Li 2 O(s) + H 2 O(l) 12) 13) The standard enthalpy of formation for CuSO 4 · 5H 2 O(s) is - 2278.0 kJ/mole at 25 ° C. The chemical equation to which this value applies

Explanation / Answer

Solution :-

10) When 1.50 g of Ba( s ) is added to 100.00 g of water in a container open to the atmosphere, the reaction shown below occurs and the temperature of the resulting solution rises from 22.00 ° C to 33.10 ° C. If the specific heat of the solution is 4.18 J/(g · ° C), calculate H for the reaction, as written. Ba( s ) + 2 H 2 O( l ) Ba(OH) 2 ( aq ) + H 2 ( g ) H = ?

Solution :-

Formula

q = m*c*delta T

q =100.00 g * 4.18 J per g C * (33.10 C – 22.00 C)

q= 4640 J

since the heat is given off so the process is exothermic so it have negative sign

(-4640 J * 137.33 g Ba / 1.500 g) (* 1 kJ / 1000 J) = -425 kJ /mol

So the delta H reaction is -425 kJ/mol

11) For the reaction 2CH 4 ( g ) + 3 Cl 2 ( g ) 2 CHCl 3 ( l ) + 3 H 2 ( g ), H ° = - 118.6 kJ. H ° f = - 134.1 kJ/mol for CHCl 3 ( l ). Find H ° f for CH 4 ( g ).

Solution :-

Delta H rxn = sum of delta Hf product – sum of delta Hf reactant

-118.6 kJ = [CHCl3*2] –[CH4*2]

-118.6 kJ = [-134.1*2] – [CH4*2]

-118.6 = -268.4 – [CH4*2]

149.8 = - 2*CH4

149.8/-2 = CH4

-74.9 kJ= CH4

Therefore the Delta Hf CH4 is -74.9 kJ/mol

12) 2 LiOH(s) --- > Li 2 O(s) + H 2 O(l) H ° = 379.1 kJ

LiH(s) + H 2 O(l) ----- > LiOH(s) + H 2 (g) H ° = - 111.0 kJ

2 H 2 (g) + O 2 (g) ----- > 2 H 2 O(l) H ° = - 285.9 kJ

Compute H ° in kJ for 2 LiH(s) + O 2 (g) Li 2 O(s) + H 2 O(l)

To get the desired equation we need to multiply equation 2 by factor 2

Then we get

2 LiOH(s) --- > Li 2 O(s) + H 2 O(l) H ° = 379.1 kJ

2 LiH(s) + 2H2O(l) ----- > 2LiOH(s) +2 H2 (g) H ° = -222.0 kJ

2 H2 (g) + O2 (g) ----- > 2 H2O(l) H ° = - 285.9 kJ

By adding these three equations by cancelling the same species on the opposite sides we get the desired equation.

2 LiH(s) + O 2 (g)----- > Li 2 O(s) + H 2 O(l)

Delta H = 379.1 kJ + (-222.0 kJ) + (-285.9 kJ) = -128.8 kJ

Therefore the delta H for the desired reaction is -128.8 kJ

13) The standard enthalpy of formation for CuSO 4 · 5H 2 O(s) is - 2278.0 kJ/mole at 25 ° C. The chemical equation to which this value applies

Chemical equation is

CuSO4(s) + 5H2O(l) ------ > CuSO4*5H2O(s)

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