Ask question

Ask question

All categories

3 years ago

8

1,4-Pentadiene has a AHhydro = -254 kJ/mol while trans-1,3-pentadiene has a AHhydra = -226 kJ/mol. Explain this difference in he

at of hydrogenation.

1 answer:

julsineya [31]3 years ago

5 0

Answer:

trans-1,3-pentadiene is more stable than 1,4-pentadiene due to presence of a conjugated double bond.

Explanation:

Here, $\Delta H_{hydro}=H(hydrogenated pdt.)-H(diene)$

H(hydrogenated pdt.) is same for both 1,4-pentadiene and 1,3-pentadiene as they both produce pentane after hydrogenation

H(diene) depends on stability of diene.

More stable a diene, lesser will be it's H(diene) value (more neagtive).

trans-1,3-pentadiene is more stable than 1,4-pentadiene due to presence of a conjugated double bond.

Hence, $\Delta H_{hydro}$ is higher (less negative) for trans-1,3-pentadiene

You might be interested in

Help !!!!!!! cell membrane plants information due soon

Leno4ka [110]

The cell membrane acts as a semi-permeable barrier in plants. In other words, is in charge or control of what comes in and out of the cell.

5 0

2 years ago

At 2°C, the vapor pressure of pure water is 23.76 mmHg and that of a certain seawater sample is 23.09 mmHg. Assuming that seawat

wariber [46]

Answer:

0.808 M

Explanation:

Using Raoult's Law

$\frac{P_s}{Pi}= x_i$

where:

$P_s$ = vapor pressure of sea water( solution) = 23.09 mmHg

$P_i$ = vapor pressure of pure water (solute) = 23.76 mmHg

$x_i$ = mole fraction of water

∴

$\frac{23.09}{23.76}= x_i$

$x_i = 0.9718$

$x_i+x_2=1$

$x_2 = 1- x_i$

$x_2 = 1- 0.9718$

$x_2 = 0.0282$

$x_i = \frac{n_i}{n_i+n_2}$ ------ equation (1)

$x_2 = \frac{n_2}{n_i+n_2}$ ------ equation (2)

where; $(n_2) =$ number of moles of sea water

$(n_i) =$ number of moles of pure water

equating above equation 1 and 2; we have :

$\frac{n_2}{n_i}$ $= \frac{0.0282}{0.9178}$

$= 0.02901$

NOW, Molarity = $\frac{moles of sea water}{mass of pure water }*1000$

$= \frac{0.02901}{18}*1000$

$= 0.001616*1000$

$= 1.616 M$

As we assume that the sea water contains only NaCl, if NaCl dissociates to Na⁺ and Cl⁻; we have $\frac{1.616}{2} =0.808 M$

3 0

2 years ago

Read 2 more answers

A 25.00 g solid sample of Ca(OH)2 was added into 1250 mL of 0.400 M HCl aqueous solution. The temperature of the solution was de

daser333 [38]

Answer:

86.735 kJ

Explanation:

Simply multiply the change in temperature by the Ccal;

(36.6 - 20.0)×5.225 = 86.735

5 0

2 years ago

I’m really bad at science and i need to know how to define it

postnew [5]

It is a column of elements in the periodic table of the chemical elements.

5 0

3 years ago

What is the volume, in mL, of 0.50 moles of ammonia (NH3) at STP?

blagie [28]

For the purpose we will here use the ideal gas law:

p×V=n×R×T

V= ?
n = 0.5 moleT= 273.15 K (at STP)
p= 101.325 kPa (at STP)
R is universal gas constant, and its value is 8.314 J/mol×K

Now when we have all necessary date we can calculate the number of moles:
V=nxRxT/p
V=0.5x8.314x273.15/101.325= 11.2 L = 11200 mL

Answer: D.

5 0

2 years ago