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3 years ago

g calculate the quantity of energy produced per gram of uranium 235 for the neutron induced Fusion of uranium

Chemistry

1 answer:

Debora [2.8K]3 years ago

7 0

Answer:

Energy released by 1 g of U-235 = 7.603 * 10¹⁰ J

Explanation:

Uranium-235, U-235 undergoes neutron-induced fission to give the following products:

1 neutron + ²³⁵U --> ¹⁴¹Ba + ⁹²Kr + 3 neutrons

Masses of reactants and products:

neutron = 1.009 amu, uranium-235 = 235.044 amu, Ba-141 = 140.910 amu, Kr-92 = 91.910 amu

mass defect = mass of reactants - mass of products

mass defect = (235.044 + 1.009) - (140.910 + 91.910 + 3 * 1.009)

mass defect = 236.053 - 235.847 = 0.206 amu

1 amu = 1.6 * 10⁻²⁷ kg

Using E = mc²

E = 0.206 * 1.6* 10⁻²⁷ kg * (3 * 10⁸ m/s)² = 2.966 * 10⁻¹¹ J

therefore, 1 atom of U-235 releases 2.966 * 10⁻¹¹ J of energy

energy released by 1 g of U-235 can be calculated as follows:

1 mole or 253 g of U-235 contains 6.02 * 10³ atoms

1 g of U-235 will contain 6.02 * 10³/235 = 2.563 * 10²¹ atoms

Energy released by 1 g of U-235 = 2.563 * 10²¹ * 2.966 * 10⁻¹¹ J = 7.603 * 10¹⁰ J

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<u>Answer:</u> The value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

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Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = 3.45

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-1=1$

Putting values in above equation, we get:

$3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084$

The equation used to calculate concentration of a solution is:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

Initial moles of $(CH_3)_3CCl(g)$ = 1.00 mol

Volume of the flask = 5.00 L

So, $\text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M$

For the given chemical reaction:

$(CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)$

Initial: 0.2 - -

At Eqllm: 0.2 - x x x

The expression of $K_c$ for above reaction follows:

$K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}$

Putting values in above equation, we get:

$0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178$

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

$[(CH_3)_2C=CH]=x=0.094M$

$[HCl]=x=0.094M$

$[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M$

Hence, the value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

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