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KatRina [158]
3 years ago
11

The reaction of hydrogen bromide(g) with chlorine(g) to form hydrogen chloride(g) and bromine(g) proceeds as follows: 2HBr(g) +

Cl2(g)2HCl(g) + Br2(g) When 23.9 g HBr(g) reacts with sufficient Cl2(g), 12.0 kJ is evolved. Calculate the value of rH for the chemical equation given.
Chemistry
1 answer:
VladimirAG [237]3 years ago
8 0

Answer:

The enthalpy of reaction per mole of HBr for this reaction = ΔrH = -40.62 kJ/mole.

Explanation:

2HBr(g) + Cl2(g) → 2HCl(g) + Br2(g)

When 23.9 g HBr(g) reacts with sufficient Cl2(g), 12.0 kJ of heat is evolved, calculate the value of ΔrH for the chemical reaction.

Note that ΔrH is the enthalpy per mole for the reaction.

Molar mass of HBr (g) = 80.91 g/mol.

Hence, 1 mole of HBr = 80.91 g

23.9 g of HBr led to the reaction giving off 12.0 kJ of heat

80.91 g of HBr will lead to the evolution of (80.91 × 12/23.9) = 40.62 kJ heat is given off.

Hence, 40.62 kJ of heat is given off per 80.91 g of HBr.

This directly translates to that 40.62 kJ of heat is given off per 1 mole of HBr

Hence, the heat given off per mole of HBr for this reaction is 40.62 kJ/mole.

But since the reaction liberates heat, it means the reaction is exothermic and the enthalpy change for the reaction (ΔHrxn) is negative.

Hence, ΔrH = -40.62 kJ/mole.

Hope this Helps!!!

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According to naming rules, the types of compound that use prefixes in their names are ________.
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Explanation:

Hello,

In this case, when forming chemical bonds in order to form compounds, we say that if electrons are shared, covalent compounds are to be formed and they usually have subscripts that need prefixes to be named, for instance phosphorous pentachloride (PCl5), dichlorine heptoxide (Cl2O7), carbon tetrachloride (CCl4) and many others.

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4 years ago
A 2.5 g sample of french fries is placed in a calorimeter with 500.0 g of water at an initial temperature of 21 °C. After combus
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<span>"What is the caloric value (kcal/g) of the french fries?"
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How many moles are in 4561 g of C3H7OH?
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3 years ago
ubstance A undergoes a first order reaction A®B with a half-life, t½, of 20 min at 25 °C. If the initial concentration of A in a
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Answer : The concentration of A after 80 min is, 0.100 M

Explanation :

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First we have to calculate the rate constant, we use the formula :

k=\frac{0.693}{t_{1/2}}

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k=3.465\times 10^{-2}\text{ min}^{-1}

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 3.465\times 10^{-2}\text{ min}^{-1}

t = time passed by the sample  = 80 min

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a - x = amount left after decay process = ?

Now put all the given values in above equation, we get

80=\frac{2.303}{3.465\times 10^{-2}}\log\frac{1.6}{a-x}

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Therefore, the concentration of A after 80 min is, 0.100 M

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