Based on the following thermochemical reactio | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) To find the value of $x$ for the reaction $C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}$,we can use Hess's Law by adding the given equations:$(1) H_2O

Vedclass · Accepted Answer

$-393 \ kJ$

Vedclass · Answer

(A) To find the value of $x$ for the reaction $C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}$,we can use Hess's Law by adding the given equations:$(1) H_2O_{(g)} + C_{(s)} \rightarrow CO_{(g)} + H_{2(g)} ; \Delta H_1 = 131 \ kJ$$(2) CO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{2(g)} ; \Delta H_2 = -282 \ kJ$$(3) H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(g)} ; \Delta H_3 = -242 \ kJ$Adding equations $(1)$,$(2)$,and $(3)$:$H_2O_{(g)} + C_{(s)} + CO_{(g)} + \frac{1}{2} O_{2(g)} + H_{2(g)} + \frac{1}{2} O_{2(g)}$ $\rightarrow CO_{(g)} + H_{2(g)} + CO_{2(g)} + H_2O_{(g)}$Canceling common terms on both sides gives:$C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}$Therefore,$x = \Delta H_1 + \Delta H_2 + \Delta H_3 = 131 + (-282) + (-242) = -393 \ kJ$.

Similar Questions

The enthalpy change for the reaction $C(s) + O_2(g) \to CO_2(g)$ is:

Given $H_{2(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)}$; $\Delta H = -44 \, Kcal$ and $2Na_{(s)} + 2HCl_{(g)} \rightarrow 2NaCl_{(s)} + H_{2(g)}$; $\Delta H = -152 \, Kcal$,calculate $\Delta H$ for the reaction $Na_{(s)} + 0.5 Cl_{2(g)} \rightarrow NaCl_{(s)}$ in $Kcal$.

The value of heat generated when $36.5 \, g$ of $HCl$ and $40 \, g$ of $NaOH$ react during neutralization is.....$kcal$.

Given two processes:
$\frac{1}{2} P_{4(s)} + 3 Cl_{2(g)} \to 2 PCl_{3(l)} \;; \Delta H = -635 \ kJ$
$PCl_{3(l)} + Cl_{2(g)} \to PCl_{5(s)} \;; \Delta H = -137 \ kJ$
The value of the heat of formation of $PCl_{5(s)}$ is ...... $kJ \ mol^{-1}$.

The enthalpy change of which reaction corresponds to $\Delta H_f^o$ for $Na_2CO_{3(s)}$ at $298 \ K$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module