Chemistry, 24.12.2019 00:31 puzzledprincess8037

Can anyone me with this ! my exam is day after tomorrow

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Answer from: orlando19882000

a. Lithium is in its standard state

b. $Li (s) + \frac{1}{2} O_2 (g) + \frac{1}{2} H_2 (g)\rightarrow LiOH (s)$

c. -440 kJ/mol

Explanation:

a. Elements in their standard states at room temperature and 1 atm pressure would have an enthalpy of formation of 0 kJ/mol. Lithium is metal at standard conditions, so its enthalpy of formation is 0 kJ/mol.

b. The equation representing the formation of a compound should following the rules below:

strictly 1 mol of the product formed;all reactants in their standard states at room temperature and 1 atm pressure.

We should, hence, form 1 mol of LiOH from the following species:

: solid lithium metal in its standard state; O_2 (g)

: oxygen gas as diatomic in its standard state; H_2 (g)

: hydrogen gas as diatomic in its standard state.

We obtain the following equation:

$Li (s) + \frac{1}{2} O_2 (g) + \frac{1}{2} H_2 (g)\rightarrow LiOH (s)$

c. Firstly, write the equation for the enthalpy of formation of water using the guidelines in (b):

$H_2 (g) + \frac{1}{2} O_2 (g)\rightarrow H_2O (l); \Delta H_1 = -286 kJ/mol$

Now given the equation:

$LiOH (s) \rightarrow Li^+ (aq) + OH^- (aq); \Delta H_2 = -21 kJ/mol$

As well as:

$Li (s) + \frac{1}{2} O_2 (g) + \frac{1}{2} H_2 (g)\rightarrow LiOH (s); \Delta H_3 = -485 kJ/mol$

Notice that multiplying reaction (2) by 2, multiplying reaction (3) by 2, multiplying reaction (1) by -2 (that is, multiplying by 2 and reversing it) and adding them together will yield the target equation:

$2 Li (s) + 2 H_2O (l)\rightarrow 2 Li^+ (aq) + 2 OH^- (aq) + H_2 (g); \Delta H_4$

According to Hess's Law, we will perform the same steps with the enthalpy values and we will add them to get the final enthalpy value:

$\Delta H_4 = 2\Delta H_2 + 2\Delta H_3 + (-2)\Delta H_1 = 2\cdot (-21 kJ/mol) + 2\cdot (-485 kJ/mol) - 2\cdot (-286 kJ/mol) = -440 kJ/mol$

Answer from: Quest

by cartels in other countries, shipped by air, mail, or land. i cant really see your answer choices : /

Answer from: Quest

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

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