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Vlad [161]
3 years ago
6

Chemical A and Chemical B react in an exothermic reaction. What can be known about what will happen when Chemical A and Chemical

B are mixed together?

Chemistry
2 answers:
rusak2 [61]3 years ago
6 0

Answer:

More energy will be released from the old substance than the new substance will need to form it's chemical bonds.

Explanation:

Given

Chemical A and Chemical B react in an exothermic reaction.

This means energy is released when they react to give products. The product formation also requires energy, which is got from the energy released when A and B reacts. The energy that is left after the formation of products is released as heat. So in order for a reaction to be exothermic, the energy released  by the reactants should be more than the energy required for the new chemical bonds to be formed to give the products.

alexandr1967 [171]3 years ago
4 0

Answer:

More energy will be released from the old substance than the new substance will need to form it's chemical bonds.

Explanation:

hi ༼ つ ◕_◕ ༽つ

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(1.7m - 2) + (2m – 4)
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Answer: 3.7 m − 6

Explanation: Simplify the expression.

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3 years ago
Coal gasification is a multistep process to convert coal into cleaner-burning fuels. In one step, a coal sample reacts with supe
ddd [48]

Answer :

The enthalpy of reaction is, -187.6 kJ/mol

The total heat will be, -2251 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

(a) The formation of CH_4 will be,

2C(coal)+2H_2O(g)\rightarrow CH_4(g)+CO_2(g)    \Delta H_{rxn}=?

The intermediate balanced chemical reaction will be,

(1) C(coal)+H_2O(g)\rightarrow CO(g)+H_2(g)     \Delta H_1=29.7kJ

(2) CO(g)+H_2O(g)\rightarrow CO_2(g)+H_2(g)    \Delta H_2=-41kJ

(3) CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)    \Delta H_3=-206kJ

We are multiplying equation 1 by 2 and then adding all the equations, we get :

(b) The expression for enthalpy of reaction will be,

\Delta H_{rxn}=2\times \Delta H_1+\Delta H_2+\Delta H_3

\Delta H_{rxn}=(2\times 29.7)+(-41)+(-206)

\Delta H_{rxn}=-187.6kJ/mol

Therefore, the enthalpy of reaction is, -187.6 kJ/mol

(c) Now we have to calculate the total heat.

\Delta H=\frac{q}{n}

or,

q=\Delta H\times n

where,

\Delta H = enthalpy change = -187.6 kJ/mol

q = heat = ?

n = number of moles of coal = \frac{1.00\times 1000g}{12.00g/mol}=83.33mol

Now put all the given values in the above formula, we get:

q=(-187.6kJ/mol)\times (83.33mol)=-2.251kJ

Thus, the total heat will be, -2251 kJ

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