1. 2M(s)+6HCl(aq) \Rightarrow 2MCl3(aq)+3H2(g)2M(s)+6HCl(aq) \Rightarrow 2MCl3(aq)+3H2(g) \triangle H_{1}=-819.0kJ 2. HCl(g) \Rightarrow HCl(aq)HCl(g)

Question

1. 2M(s)+6HCl(aq) \Rightarrow 2MCl3(aq)+3H2(g)2M(s)+6HCl(aq) \Rightarrow 2MCl3(aq)+3H2(g) \triangle H_{1}=-819.0kJ 2. HCl(g) \Rightarrow HCl(aq)HCl(g)

Jean Blumer 2022-01-10 Answered

1.

2 M (s) + 6 H C l (a q) \Rightarrow 2 M C l 3 (a q) + 3 H 2 (g) 2 M (s) + 6 H C l (a q) \Rightarrow 2 M C l 3 (a q) + 3 H 2 (g)

△ H_{1} = - 819.0 k J

2.

H C l (g) \Rightarrow H C l (a q) H C l (g) \Rightarrow H C l (a q)

△ H_{2} = - 74.8 k J

3.

H 2 (g) + C l 2 (g) \Rightarrow 2 H C l (g) H 2 (g) + C l 2 (g) \Rightarrow 2 H C l (g)

△ H_{3} = - 1845.0 k J

4.

M C l 3 (s) \Rightarrow M C l 3 (a q)

△ H_{4} = - 128.0 k J

Use the given information to determine the enthalpy of the reaction

2 M (s) + 3 C l_{2} (g) \Rightarrow 2 M C l_{3} (s)

△ H = ?

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Expert Answer

Melinda McCombs

Answered 2022-01-11 Author has 38 answers

Step 1
In the above reactions we only need 2 reactions to calculate the enthalpy of the following reaction;

2 M (s) + 3 C l 2 (g) \Rightarrow M C l 3 (g)

Step 2
We have 4 chemical equations from which 2 are important:

2 M (s) + 6 H C l (a q) \Rightarrow 2 M C l_{3} (a q) + 3 H_{2} (g)

△ H_{1} = 819 K J

H_{2} (g) + C l_{2} (g) \Rightarrow 2 H C l (g)

△ H_{3} = - 1845 K J

Multiplying equation 1st by 3

3 H_{2} (g) + 3 C l_{2} (g) \Rightarrow 6 H C l (g)

△ H_{3} = 3 \times (- 1845 K J) = - 5535 K J

Adding equation 1st and 3rd

2 M (s) + 3 C l_{2} (g) \Rightarrow 2 M C l_{3} (a q)

△ H_{r e a c t i o n}

= (△ H_{1} + △ H_{3}) = - 4716 K J

Therefore the enthalpy of formation

= - 4716 K J

(answer)

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Ben Owens

Answered 2022-01-12 Author has 27 answers

Chemical reaction 1:

2 M (s) + 6 H C l (a q) \Rightarrow 2 M C l 3 (a q) + 3 H 2 (g); △ H 1 = - 556.0 k J

. Chemical reaction 2:

H C l (g) \Rightarrow H C l (a q); △ H 2 = - 74.8 k J

.
Chemical reaction 3:

H 2 (g) + C l 2 (g) \Rightarrow 2 H C l (g); △ H 3 = - 1845.0 k J

Chemical reaction 4:

M C l 3 (s) \Rightarrow M C l 3 (a q); △ H 4 = - 342.0 k J

.
Chemical reaction 5:

2 M (s) + 3 C l 2 (g) \Rightarrow 2 M C l 3 (s); △ H 5 = ? . △ H 5 = △ H 1 + 6 \cdot △ H 2 + 3 \cdot △ H 3 - 2 \cdot △ H 4

.

△ H 5 = - 550 k J + 6 \cdot (- 74, 8 k J) + 3 \cdot (- 1845 k J) - 2 \cdot (- 342 k J)

.

△ H 5 = - 550 k J - 448, 8 k J - 5535 k J + 684 k J

.

△ H 5 = - 5849, 8 k J

.

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star233

Answered 2022-01-15 Author has 137 answers

On adding,
$2 M (s) + 3 C l_{2} (g) \Rightarrow 2 M C l_{3} (s) △ H = - 944 k J + 6 \times (- 74.8 k J) + 3 \times (- 1845 k J) + (2) \times (234 k J) △ H = - 944 k J - 448.8 - 5535 k J + 468 k J △ H = - 6459.8 k J$

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