Problem 2 Calculate the standard enthalpy of formation (forming one mole compoun

Question

Problem 2 Calculate the standard enthalpy of formation (forming one mole compound from its elements in their standard states) of solid calcium hydroxide. Showing all work includes rewriting the given thermodynamic equations to get the target equation AND your final calculation for the standard enthalpy of formation. Give the target equation (balanced chemical equation with states of matter for the standard enthalpy of formation for calcium hydroxide) Given thermodynamic equations: &H;=-431.2 k] . mol- &H;° =-1624 kj . mor1 &H;=-285.8 kJ-mol-1 Ca(OH)2(s) Ca(a) + 20Haq) · 112(g) +302(g) H2O(1)

Explanation / Answer

Problem 2

The first equation given is not correctly balanced equation

It is

Ca + 2H2O -----------> Ca+2 + 2OH- + H2

The required equation is

Ca(s) + H2(g) + O2(g) ----------> Ca(OH)2(s)

To get this equation from the given equations we have to

eq(1) - equation 2 + [2 x equation 3 ]

thus delta H for required equation = delta H1 - delta H2 + 2 x delta H3

= -431.2 -(-16.24) +2 (-285.8) kJ

= -970.32 kJ

Probelm 2

Among the given transitions

a) n=2 to n=4

n=1 to n=2

transitions occur with no emission of photon , as they absorb photons.

b) the transition n=2 to n=1 results smaller energy emission among the given emissions , thus has longest wavelength.

c) transition n=4 to n=1 involves largest energy photon emission thus have largest frequency.

d) the emission n=2 to n=1 requires smaller energy , thus smaller frequency.

QB) I take the last electron of Ga which has 4p1 configuration.

The set of quantum numbers are

n=4

l=1

ml = +1

ms= +1/2

C) In the 3 shell of Thallium , the configuration is 3s23p63d10

so for an electron in 3s , the set of quantum numbers are

n=3, l =0, m= 0 and ms = +1/2 or -1/2

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