Question

Calculate the specific heat of titanium metal in (J/g-degreeC) if it takes 89.8 J to increase the temperature of 33.21 g of this metal by 5.30 degreeC. Show formula, rearranged formula, then substitute and calculate Given two reactions: 2Cu (s) + O_2 (g) rightarrow 2CuO (s) delta H^0 = -310 kJ 2Cu (s) + 1/2 O_2 (g) rightarrow Cu_2O (s) delta H^0 = -169 kJ Calculate value of delta H^0 for the following reaction by using the above reactions. by writing reactions again and all calculation. If not shown no points given. Cu_2 O (s) + 1/2 O_2 (g) rightarrow 2CuO (s) deltaH^0 =

Answer 1

7. We know that basic formula to solve these type of problems is :

$Q = c \times m\times \bigtriangleup T$

$Where, Q = heat\ added, \ c = specific \ heat , m= mass \ and \bigtriangleup T = change \ in \ temperature$

Rearranging formula we get :

$Q = c \times m\times \bigtriangleup T$

$c = \frac{Q}{m\times \bigtriangleup T}$

Substituting values :

$c = \frac{Q}{m\times \bigtriangleup T}$

$c = \frac{89.8}{33.21\times 5.30}$

$c = 0.5101\ \ \frac{J}{g\ ^{0}C}$

8.

2 Cu + O₂ -----------> 2 CuO ...............(1) $\bigtriangleup H_{1} = -310\ KJ$

2 Cu + (1/2) O₂ -----------> Cu₂O...........(2)    $\bigtriangleup H_{2} = -169\ KJ$

Reversing the reaction (2) we get,

Cu₂O-----------> 2 Cu + (1/2) O₂ ........(3) $\bigtriangleup H_{3} = 169\ KJ$

Now adding reaction (1) and (3) we get,

2 Cu + O₂ -----------> 2 CuO .......................(1)    $\bigtriangleup H_{1} = -310\ KJ$

+ Cu₂O -----------> 2 Cu + (1/2) O₂ ..........(3) + $\bigtriangleup H_{3} = 169\ KJ$

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2 Cu + O₂ + Cu₂O -----------------> 2 CuO + 2 Cu + (1/2) O₂    $\bigtriangleup H_{net} = -310 + 169\ KJ$

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Simplifying above equation and removing common compounds from both sides, we get

Cu₂O + O₂ -----------------> 2 CuO + (1/2) O₂ $\bigtriangleup H_{net} = -141\ KJ$

Further simplifying ,we get

Cu₂O + O₂ - (1/2) O₂ -----------------> 2 CuO $\bigtriangleup H_{net} = -141\ KJ$

Cu₂O + (1/2) O₂ -----------------> 2 CuO $\bigtriangleup H_{net} = -141\ KJ$