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1 year ago

How do chemical and nuclear reactions differ in(c) Effect on rate of higher reactant concentration?

1 answer:

7 0

Nuclear reactions involve a change in an atom's nucleus, usually producing a different element. Chemical reactions, on the other hand, involve only a rearrangement of electrons and do not involve changes in the nuclei.

<h3>What affects the rate of nuclear reactions?</h3>

Reactant concentration, the physical state of the reactants, and surface area, temperature, and the presence of a catalyst are the four main factors that affect reaction rate.

<h3>What is the main difference between chemical reactions and nuclear reactions?</h3>

Chemical reaction normally occurs outside the nucleus. Nuclear reaction happens only inside the nucleus. When chemical reactions occur elements hold their identity and the nuclei of atoms also remains unchanged. During nuclear reactions, the nuclei of atoms changes completely and new elements are formed.

Learn more about chemical reaction here:

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3 years ago

What does an atom become when it gains an electron in an ionic bond?

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3 years ago

Consider the reaction between iodine gas and chlorine gas to form iodine monochloride. A reaction mixture at 298.15k initially c

uranmaximum [27]

Step 1 : Write balanced chemical equation

The balanced chemical equation for the reaction between iodine gas and chlorine gas is given below.

$I_{2} (g) + Cl_{2} (g) -----> 2 ICl (g)$

Step 2 : Set up ICE table

We will set up an ICE table for the above reaction

Following points are considered while drawing ICE table

- The initial concentration of product is assumed as 0

- The change in concentration (C) is assumed as x. Change (x) is negative for reactants and positive for products

- The coefficients in balanced equation are considered while writing C values

Check attached file for ICE table

Step 3 : Set up equilibrium constant equation

The equation for equilibrium constant can be written as

$K_{eq} = \frac{[ICl]^{2}}{[I_{2}][Cl_{2}]}$

Step 4 : Solving for x

Keq at 298.15 K is given as 81.9

Let us plug in the equilibrium values (E) for I₂, Cl₂ and ICl from ICE table

81.9 = $\frac{(2x)^{2}}{(0.437-x) (0.269-x)}$

81.9 = $\frac{(2x)^{2}}{x^{2} -0.706x + 0.118}$

$(2x)^{2} = 81.9 [ x^{2} -0.706x + 0.118]$

$4x^{2} = 81.9x^{2} -57.8x +9.66$

$77.9x^{2} -57.8x+9.66 = 0$

Solving the above equation using quadratic formula we get

x = 0.488 or x = 0.254

The value 0.488 cannot be used because the change (C) cannot be greater that initial concentration of the reactants.

Therefore the change in concentration of the gases during the reaction is 0.254 M

Hence, x = 0.254 M

From the ICE table, we know that the equilibrium concentration of ICl is 2x

$[ICl]_{eq} = 2 ( 0.254) = 0.508 M$

The concentration of ICl when the reaction reaches equilibrium is 0.508 M

6 0

3 years ago

Use the information provided to determine ΔH°rxn for the following reaction: ΔH°f (kJ/mol) 3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2

zepelin [54]

Answer:

ΔH°rxn = -47 kJ

Explanation:

Using Hess´s law for the reaction:

3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g) ,

the ΔH°rxn will be given by the expression:

ΔH°rxn kJ = 2ΔHºf(Fe3O4) + ΔHºf(CO2) - ( 3ΔHºf(Fe2O3) + ΔHºf(CO) )

= 2(-1118) + (-394) - ( 3( -824 ) + ( -111 ) )

= - 47 kJ

3 0

3 years ago

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