Science, asked by khush20101, 3 months ago

Given:
C + 2S → CS2; AH = +117 kJ mol-1
C + 02 → CO2; AH° = -393 kJ mol-1
S+02 S02; AH = -297 kJ mol-1
The value of heat of combustion of CS2 in kJ mol-1 is :-​

Answers

Answered by javeediqbalkgn
0

Answer:

BrainlyHelper

BrainlyHelper

27.12.2017

Chemistry

Secondary School

Question 6.19 For the reaction

2A(g) + B(g)→2D(g)

ΔUθ = –10.5 kJ and ΔSθ= –44.1 JK–1.

Calculate ΔGθ for the reaction, and predict whether the reaction may occur spontaneously.

Class XI Thermodynamics Page 183

2

SEE ANSWERS

abhi178

abhi178

2A(g) + B(g) -------> 2D(g) : ΔU° = -10.5 KJ

Δng = np - nr

= 2 - (2 + 1) = 2 - 3 = -1

ΔH° = ΔU° + ΔΔngRT, from formula,

= -10.5 KJ + (-1) × 0.008314 KJ/K/mol × 298K

=-10.5 KJ - 2.447 KJ

= -12.947 KJ

now, use the formula, ΔG° = ΔH° - TΔS°

= -12.947 KJ - 298K × (-44 × 10⁻³ J/K)

= -12.947 KJ + 13.14 KJ

= +0.165 KJ

the reaction will not occur spontanteously because ΔG° is positive.

Brainly User

Hi

Here is your answer,

2A(g) + B(g) → 2D ; Δng = np - nr = 2-3 = -1

ΔH° = ΔU° + ΔngRT

ΔH° = -10.5 kJ + ( -1 × 8.314 × 10⁻³ kJ mol⁻¹ × 298K )

ΔH° = -10.5 + (- 2.477) kJ mol⁻¹

ΔH°= -12.977 kJ mol⁻¹¹

ΔG° = ΔH° - TΔS°

ΔG° = -12.977 kJ mol⁻¹ - (298 K × -44.1 × 10⁻³ kJ⁻¹ mol⁻¹)

ΔG° = -12.977 kJ mol⁻¹ + 13.14 kJ mol⁻¹ = + 0.165 kJ mol⁻¹

∴ here reaction will not occur spontaneously because ΔG° is positive.

Hope it helps you !

Explanation:

Mark as BRAINLIST

Similar questions