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lilavasa [31]
3 years ago
9

Be sure to answer all parts. The thermal decomposition of phosphine (PH3) into phosphorus and molecular hydrogen is a first-orde

r reaction: 4PH3(g) → P4(g) + 6H2(g) The half-life of the reaction is 35.0 s at 680°C.
Chemistry
1 answer:
amm18123 years ago
5 0

The given question is incomplete. The complete question is :

The thermal decomposition of phosphine (PH3) into phosphorus and molecular hydrogen is a first-order reaction: 4PH_3(g)\rightarrow P_4(g)+6H_2(g) The half-life of the reaction is 35.0 s at 680°C. Calculate the first order rate constant.

Answer: a) The first order rate constant is 0.0198s^{-1}

b) The time after which 95% reactions gets completed is 151 seconds

Explanation:

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

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process  

a) for finding the rate constant

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

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

k=\frac{0.693}{35.0s}=0.0198s^{-1}

The first order rate constant is 0.0198s^{-1}

b) for completion of 95 % of reaction

t=\frac{2.303}{k}\log\frac{100}{100-95}

t=\frac{2.303}{0.0198}\log\frac{100}{5}

t=151 s

The time after which 95% reactions gets completed is 151 seconds

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The enthalpy of formation of MX is ΔHf° = –525 kJ/mol. The enthalpy of sublimation of M is ΔHsub = 139 kJ/mol. The ionization en
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3 0
2 years ago
A 12.5 g sample of granite initially at 82.0 oc is immersed into 25.0 g of water that is initially at 22.0 oc. what is the final
Paraphin [41]

The amount of heat lost by granite is equal to the amount of heat gained by water. Therefore their change in enthalpies must be equal. The opposite in sign means that one is gaining while the other is losing

ΔH granite = - ΔH water

ΔH is the change in enthalpy experienced by a closed object as it undergoes change in energy. This is expressed mathematically as,

ΔH = m Cp (T2 – T1)

Given this information, we can say that:

12.5 g * 0.790 J / g ˚C * (T2 – 82 ˚C) = - 25.0 g * 4.18 J / g ˚C * (T2 – 22 ˚C)

9.875 (T2 – 82) = 104.5 (22 – T2)

9.875 T2 – 809.75 = 2299 – 104.5 T2

114.375 T2 = 3108.75

T2 = 27.18 ˚C

The temperature of 2 objects after reaching thermal equilibrium is 27.18 ˚<span>C.</span>

7 0
3 years ago
Based on the equation, how many grams of Br2 are required to react completely with 36.2 grams of AlCl3?
s2008m [1.1K]

Answer:

65.08 g.

Explanation:

  • For the reaction, the balanced equation is:

<em>2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,</em>

2.0 mole of AlCl₃ reacts with 3.0 mole of Br₂ to produce 2.0 mole of AlBr₃ and 3.0 mole of Cl₂.

  • Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:

<em>n = mass/molar mass</em> = (36.2 g)/(133.34 g/mol) = <em>0.2715 mol.</em>

<u><em>Using cross multiplication:</em></u>

2.0 mole of AlCl₃ reacts with → 3.0 mole of Br₂, from the stichiometry.

0.2715 mol of AlCl₃ reacts with → ??? mole of Br₂.

∴ The no. of moles of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃  = (0.2715 mol)(3.0 mole)/(2.0 mole) = 0.4072 mol.

<em>∴ The mass of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = no. of moles of Br₂ x molar mass</em> = (0.4072 mol)(159.808 g/mol ) = <em>65.08 g.</em>

4 0
3 years ago
Oxygen gas is collected over water. The total pressure (the O2 pressure the water vapor pressure) is 736 torr. The temperature o
tigry1 [53]

Answer:

The value of the partial pressure of the oxygen  P_{O_{2} } =  690 torr

Explanation:

Total pressure of the mixture of gases = 736 torr

The partial pressure of water vapor = 46 torr

From the law of pressure we know that

Total pressure = The partial pressure of water vapor + The partial pressure of oxygen O_{2}

Put the values of pressures in above equation we get,

⇒ 736 = 46 + P_{O_{2} }

⇒ P_{O_{2} } = 736 - 46

⇒ P_{O_{2} } =  690 torr

This is the value of the partial pressure of the oxygen.

3 0
3 years ago
Complete combustion of 3.20g of a hydrocarbon produced 9.69g of CO2 and 4.96g of H2O. What is the empirical formula for the hydr
vlabodo [156]

Let empirical formula for hydrocarbon is CxHy

it will undergo combustion as

CxHy + (x + y/4) O2  ---> xCO2 + (y/2 )H2O

Given that mass of CO2 produced = 9.69 g

So moles of CO2 produced = 9.69 / 44 = 0.22 moles

So moles of carbon present = 0.22 moles

mass of H2O produced = 4.96 g

Moles of H2O produced = mass / molar mass = 4.96 / 18 = 0.28 moles

So moles of H present = 2 X 0.28 = 0.56 moles

Let us divided the moles of each with lowest value of moles

Moles of Carbon = 0.22 / 0.22 = 1 moles

moles of H = 0.56 / 0.22 = 2.55

Multiplying with two to get whole number

the ratio of carbon and hydrogen will be : C:H = 2:5

empirical formula : C2H5


4 0
3 years ago
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